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# Antenna Fundamentals

An antenna is a device for converting electromagnetic radiation in space into electrical currents in conductors or vice-versa, depending on whether it is being used for receiving or for transmitting, respectively.  Passive radio telescopes are receiving antennas.  It is usually easier to calculate the properties of transmitting antennas. Fortunately, most characteristics of a transmitting antenna (e.g., its radiation pattern) are unchanged when the antenna is used for receiving, so we often use the analysis of a transmitting antenna to understand a receiving antenna used in radio astronomy.

### Radiation from a Short Dipole Antenna (Hertz Dipole)

The coordinate system used to describe the radiation from a short (total length $l \ll \lambda$) dipole driven by a current source of frequency $\nu$.

The simplest antenna is a short (total length $l$ much smaller than one wavelength $\lambda$) dipole antenna, which is shown above as two colinear conductors (e.g., wires or conducting rods).  Since they are driven at the small gap between them by a current source (a transmitter), the current in the bottom conductor is 180 deg out of phase with the current in the top conductor.  The radiation from a dipole depends on frequency, so we consider a driving current $I$ varying sinusoidally with angular frequency $\omega = 2 \pi \nu$: $$I = I_0 \cos(\omega t)~,$$ where $I_0$ is the peak current going into each half of the dipole.  It is computationally convenient to replace the trigonometric function $\cos(\omega t)$ with its exponential equivalent, the real part of
$$e^{-i \omega t} = \cos(\omega t) - i \sin(\omega t)$$ so the driving current can be rewritten as $$I = I_0 e^{-i \omega t}$$ with the implicit understanding that only the real part of $I$ represents this current. The driving current accelerates charges in the antenna conductors, so we can use Larmor's formula to calculate the radiation from the antenna by converting from the language of charges and accelerations to time-varying currents.

Recall that electric current is defined as the time derivative of electric charge:
$$I \equiv {d q \over d t}~.$$ Along a wire, the current is the amount of charge flowing past any point per unit time. For a wire on the $z$-axis
$$I = {d q \over d t} = {d q \over d z} {d z \over d t} = {d q \over d z} v~,$$ where $v$ is the instantaneous flow velocity of the charges.

It is a common misconception to believe that the velocities $v$ of individual electrons in a wire are comparable with the speed of light $c$ because electrical signals do travel down wires at nearly the speed of light.  A wire filled with electrons is like a garden hose already filled with an incompressible fluid—water.  When the faucet is turned on, water flows from the other end of a full hose almost immediately, even though individual water molecules are moving slowly along the hose.  As a specific example, consider a current of 1 ampere flowing through a copper wire of cross section $\sigma = 1$ mm$^2 = 10^{-6}$ m$^2$.  The number density of free electrons is about equal to the number density of copper atoms in the wire, $n \approx 10^{29}$ m$^{-3}$.  In mks units, the charge of an electron is
$$-e \approx 4.80 \times 10^{-12} {\rm ~statcoul} \times {1 {\rm ~coul} \over 3 \times 10^9 {\rm ~statcoul}} \approx 1.60 \times 10^{-19} {\rm ~coul}$$ One Ampere is one Coulomb per second, so the number $N$ of electrons flowing past any point along the wire in one second is
$$N = {I \over \vert e \vert} \approx {1 {\rm ~coul~s}^{-1} \over 1.60 \times 10^{-19}{\rm ~coul}} \approx 6.25 \times 10^{18}{\rm ~s}^{-1}$$ The average electron velocity is only
$$v \approx {N \over \sigma n} \approx { 6.25 \times 10^{18}{\rm ~s}^{-1} \over 10^{-6}{\rm ~m}^2 \times 10^{29}{\rm ~m}^{-3}} \approx 6 \times 10^{-5} {\rm ~m~s}^{-1} \ll c$$ Thus the nonrelativistic Larmor equation may be used directly to calculate the radiation from a wire.

From the derivation of Larmor's formula, recall that
$$E_\bot = {q \dot{v} \sin\theta \over r c^2}$$ so in the short dipole
$$E_\bot = \int_{z = -l/2}^{+l/2} {d q \over d z} d z {\dot{v} \sin\theta \over r c^2}$$ For a sinusoidal driving current,
$$\dot{v} = -i \omega v$$ and
$$E_\bot = {-i \omega \sin\theta \over r c^2} \int_{-l/2}^{+l/2} {d q \over d z} v dz$$
$$E_\bot = {-i \omega \sin\theta \over r c^2} \int_{-l/2}^{+l/2} I d z$$ That is, the radiated electric field strength $E_\bot$ is proportional to the integral of the current distribution along the antenna. The current at the center is just the driving current $I = I_0 e^{-i \omega t}$ and the current must drop to zero at the ends of the antenna, where the conductivity goes to zero. For a short antenna, we can make the approximation that the current declines linearly from the driving current at the center to zero at the ends:
$$I(z) = I_0 e^{- i \omega t} \biggl[1 - {\vert z \vert \over (l / 2)} \biggr]$$ Then
$$\int_{-l/2}^{+l/2} I dz = {I_0 l \over 2} e^{-i \omega t}$$ and
$$E_\bot = {-i \omega \sin\theta \over r c^2} {I_0 l \over 2} e^{-i \omega t}$$ Substituting $\omega = 2 \pi c / \lambda$ gives
$$E_\bot = {-i 2 \pi c \sin\theta \over \lambda r c^2 } {I_0 l \over 2} e^{-i \omega t}$$ $$E_\bot = {-i \pi \sin\theta \over c} {I_0 l \over \lambda} {e^{ -i \omega t} \over r}$$ Since $\vert\vec{E_\bot}\vert = \vert\vec{H_\bot}\vert$ (cgs), the time-averaged Poynting flux (power per unit area) is
$$\langle{S}\rangle = {c \over 4 \pi} \langle E_\bot^2 \rangle ,$$ and
$$\langle{S}\rangle = {c \over 4 \pi} \biggl({I_0 l \over \lambda} \, {\pi \over c} \biggr)^2 {\sin^2\theta \over r^2} \biggl( {1 \over 2}\biggr)$$ because $\langle \cos^2 (\omega t) \rangle = 1/2$. Note that the radiation from a short dipole has the same polarization and the same doughnut-shaped power pattern (the power pattern is the angular distribution of radiated power, often normalized to unity at the peak)
$$\bbox[border:3px blue solid,7pt]{P \propto \sin^2\theta}\rlap{\quad \rm {(3A1)}}$$ as Larmor radiation from an accelerated charge because all of the charges in the dipole are being accelerated along one line much shorter than one wavelength.  From the observer's point of view, the power received depends only on the projected (perpendicular to the line of sight) length $(l \sin\theta)$ of the dipole.  Thus the electric field received is proportional to the apparent length of the dipole. The time-averaged total power emitted is obtained by integrating the Poynting flux over the surface area of a sphere of any radius $r \gg l$ centered on the antenna:
$$\langle P \rangle = \int\langle S \rangle dA = {c \over 4 \pi} \biggl({I_0 l \over \lambda} \, {\pi \over c} \biggr)^2 \biggl({1 \over 2}\biggr) \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi {\sin^2\theta \over r^2} r \sin\theta d \phi r d\theta$$ $$\langle P \rangle = {c \over 4 \pi} \biggl({I_0 l \over \lambda} \, {\pi \over c} \biggr)^2 \biggl({1 \over 2}\biggr) 2 \pi \int_{\theta = 0}^\pi \sin^3\theta d\theta$$ Recall that $\int_0^\pi \sin^3\theta d\theta = 4/3$, so the time-averaged power radiated by a short dipole is
$$\bbox[border:3px blue solid,7pt]{\langle P \rangle = {\pi^2 \over 3 c} \biggl({I_0 l \over \lambda}\biggr)^2 }\rlap{\quad \rm {(3A2)}}~,$$ where $I_0 \cos (\omega t)$ is the driving current, $l$ is the total length of the dipole, and $\lambda$ is the wavelength.

Most practical dipoles are half-wave dipoles ($l \approx \lambda/2$) because half-wave dipoles are resonant, meaning that they provide a nearly resistive load to the transmitter.  When each half of the dipole is $\lambda/4$ long, the standing-wave current is highest at the center and naturally falls as $\cos(2 \pi z / \lambda)$ to almost zero at the ends of the conductors.

The ground-plane vertical shown below is very similar to the dipole.  A ground-plane vertical is one half of a dipole above a conducting plane, which is called a "ground plane" because historically the conducting plane for vertical antennas was the surface of the Earth.  The transmitter is connected between the base of the vertical, which is insulated from the ground, and the ground plane near the base.  Many AM broadcast transmitting antennas are tall (at $\nu \sim 1$ MHz, $\lambda \sim 300$ m and a $\lambda/4$ vertical antenna is about 75 m high) towers acting as quarter-wave verticals. The conducting ground plane is a mirror that creates the lower half of the dipole as the mirror image of the upper half.  Electric fields produced by the vertical must induce currents in the conducting plane that ensure that the horizontal component of the electric field goes to zero on the conductor.  This implies that the virtual electric fields from the image vertical must have the same amplitude but be 180 degrees out of phase, exactly as in a half-wave dipole.  Consequently the radiation field from a ground-plane vertical is identical to that of a dipole in the half space above the ground plane and zero below the ground plane.

The ground-plane vertical is just half of a dipole above a conducting plane. The lower half of the dipole is the reflection of the vertical in the mirror provided by the conducting "ground plane."  The image vertical is 180 deg out of phase with the real vertical.  Above the ground plane, the radiation from the ground-plane vertical is exactly the same as the radiation from the dipole.

According to the strict definition of an antenna as a device for converting between electromagnetic waves in space and currents in conductors, the only antennas in most radio telescopes are half-wave dipoles and their relatives, quarter-wave ground-plane verticals. The large parabolic reflector of a radio telescope serves only to focus plane waves onto the feed antenna.  [The term "feed" comes from radar antennas used for transmitting; the "feed" feeds transmitter power to the main reflector.  Receiving antennas used in radio astronomy work the other way around, and the "feed" actually collects radiation from the reflector.]

Actual half-wave dipoles, backed by small reflectors about $\lambda/4$ behind them to focus the dipole pattern in the direction of the main dish, are normally used as feeds at low frequencies ($\nu < 1$ GHz) or long wavelengths ($\lambda > 0.3$ m) because of their relatively small size.  However, the radiation patterns of  half-wave dipoles backed by small reflectors are not well matched to most parabolic dishes, so their performance is less than optimum.

At shorter wavelengths, almost all radio-telescope feeds are quarter-wave ground-plane verticals inside waveguide horns. Radiation entering the relatively large (size $> \lambda$) rectangular or circular aperture of the tapered horn is concentrated into a rectangular or circular waveguide with parallel conducting walls.  In the case of the rectangular waveguide shown below, the side walls are separated by slightly over $\lambda/2$ so that vertical electric fields can travel down the waveguide with low loss.  The top and bottom walls are separated by somewhat less than $\lambda/2$ so only one mode with vertical electric fields can propagate.  The $\lambda/4$ vertical inserted through a small hole in the bottom wall collects most of this vertically polarized radiation and converts it into an electric current that travels down the coaxial cable to the receiver.  The backshort wall about $\lambda/4$ behind the dipole ensures that the dipole sees only radiation coming from the direction of the horn opening.

Most high-frequency feeds are quarter-wave ground-plane verticals inside waveguide horns. The only true antenna in this figure is the $\lambda/4$ ground-plane vertical, which converts electromagnetic waves in the waveguide to currents in the coaxial cable extending down from the waveguide.

The power flowing through a circuit is $P = V \times I$, where $V$ is the voltage (defined as energy per unit charge) and $I$ is the current (defined as charge flow per unit time), so $P$ has dimensions of energy per unit time.  The physicist George Simon Ohm observed that the current flowing through most materials is proportional to the applied voltage, so many (but not all) objects have a well-defined resistance defined by $R = V / I$ (Ohm's law).  For them, $P = V \times I = I^2 R = V^2 / R$.

From Ohm's law for time-varying currents,
$$\langle P \rangle = \langle I^2 \rangle R$$ If $I = I_0 \cos(\omega t)$,
$$\langle P \rangle = {I_0^2 R \over 2}$$ The radiation resistance of an antenna is defined by
$$\bbox[border:3px blue solid,7pt]{R \equiv {2 \langle P \rangle \over I_0^2}}\rlap{\quad \rm {(3A3)}}$$ For our short dipole, the radiation resistance is
$$R = {2 \pi^2 \over 3 c} \biggl( {l \over \lambda} \biggr)^2$$

Example: A "half-wave" dipole has length $l = \lambda / 2$. This is the length of a resonant dipole antenna.  Resonant antennas are used in most real applications because the impedance of a resonant antenna is resistive; nonresonant antennas have large capacitive or inductive reactances as well. A half-wave dipole doesn't strictly satisfy our criterion $l \ll \lambda$ for being "short," so the current distribution along the dipole is actually closer to sinusoidal than linear and our calculated radiation resistance will not be exact. Proceeding nonetheless to estimate the radiation resistance of a half-wave dipole,
$$R \approx {2 \pi^2 \over 3\times 3\times 10^{10}{\rm ~cm~s} ^{-1}} \biggl({1 \over 2}\biggr)^2 \approx 5.5 \times 10^{-11}{\rm ~s~cm}^{-1}$$ Engineers and real test instruments use the mks "Ohm" (symbol $\Omega$) as the unit of resistance. The conversion factor is 1 $\Omega$ = $(10^{-11} / 9)$ s cm$^{-1}$, so
$$R \approx 5.5 \times 10^{-11}{{\rm s} \over {\rm cm}} {9 {\rm ~cm} \over 10^{-11} {\rm ~s}} ~\Omega \approx 50 ~\Omega$$ [The actual radiation resistance of a half-wave dipole in free space is about 73 $\Omega$.]

For a given driving current, a ground-plane vertical of height $l/2$ emits exactly like a dipole of length $l$ above the ground plane and zero below the ground plane.  Thus the total power emitted by the vertical is half the power emitted by the dipole, and the radiation resistance of the vertical is half the radiation resistance of the dipole.

The radiation resistance $R_0$ of free space can be obtained from the relations
$$\vert \vec{S} \vert = {c \over 4 \pi} E^2 {\rm ~\quad and \quad~} P = {V^2 \over R}~.$$
Since the electric field $E$ is just the voltage per unit length $V / l$ and the flux is the power per unit area $l^2$,
$$\vert \vec{S} \vert = { c V^2 \over 4 \pi l^2} = {V^2 \over R_0 l^2}$$
$$R_0 = {4 \pi \over c} = {4 \pi \over 3 \times 10^{10} {\rm ~cm~s}^{-1}}$$
Converting to mks units yields the radiation resistance of space in Ohms:
$$R_0 = {4 \pi \over 3 \times 10^{10} {\rm ~cm~s}^{-1} \times {1/9}\times 10^{-11} {\rm ~sec~cm}^{-1}~\Omega^{-1}} = 120 \pi ~\Omega \approx 377 ~\Omega~.$$

Since a black hole is a perfect absorber of radiation, its impedance must also be $120 \pi ~\Omega$ to match that of free space.  A black hole spinning in an external magnetic field can generate electrical power with a voltage/current ratio of $120 \pi ~\Omega$, and this process may be important in powering quasars (see Blandford & Znajek 1977, MNRAS, 179, 433).

The Power Gain of a Transmitting Antenna

The power gain $G(\theta,\,\phi)$ of a transmitting antenna is defined as the power transmitted per unit solid angle in direction $(\theta,\,\phi)$ divided by the power transmitted per unit solid angle from an isotropic antenna driven by a transmitter supplying the same total power. Frequently the value of $G$ is expressed logarithmically in units of decibels (dB):
$$G({\rm dB}) \equiv 10 \times \log_{10}(G)$$

For a lossless antenna, energy conservation requires that the gain averaged over all directions be
$$\langle G \rangle \equiv {\int_{\rm sphere} G d \Omega \over \int_{\rm sphere } d \Omega} = {4 \pi \over 4 \pi}$$ or $$\bbox[border:3px blue solid,7pt]{\langle G \rangle = 1}\rlap{\quad \rm {(3A4)}}$$ Consequently
$$\int_{\rm sphere} G d \Omega = \int_{\rm sphere} 1 d \Omega = 4 \pi~$$ for any lossless antenna. Different lossless antennas may radiate with different directional patterns, but they cannot alter the total amount of power radiated. Consequently, the gain of a lossless antenna depends only on the angular distribution of radiation from that antenna. In general, an antenna having peak gain $G_{\rm max}$ must beam most of its power into a solid angle $\Delta \Omega$ such that
$$\Delta \Omega \approx {4 \pi \over G _{\rm max}}~.$$ Thus the higher the gain, the narrower the beam or power pattern.

Example: What is the power gain of a short dipole? It is sufficient to recall only the angular dependence of the short-dipole power pattern $$\langle S \rangle \propto \sin^2\theta~,$$ where $\theta$ is the angle from the dipole axis.  Thus $$G \propto \sin^2\theta = G_0 \sin^2\theta~.$$ The maximum gain $G_0$ is determined by energy conservation:
$$\int_{\rm sphere} G d\Omega = \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi G_0 \sin^2\theta d\theta \sin\theta d\phi = 4 \pi~.$$ $$2 \pi G_0 \int_0^\pi \sin^3\theta d \theta = 4 \pi ~.$$ Recall that $\int_0^\pi \sin^3\theta d \theta = 4/3$ so
$$G_0 = {4 \pi \over 2 \pi}{3 \over 4} = {3 \over 2}$$ and
$$G(\theta, \phi) = {3 \sin^2\theta \over 2} ~.$$ Expressed in dB, the maximum gain $G_0$ of a short dipole is
$$G_0 = 10 \log_{10}(3/2) \approx 1.76{\rm ~dB}.$$

Note that $G(\theta, \phi)$ is independent of the antenna length in wavelengths so long as $l \ll \lambda$ because the short-dipole power pattern is independent of $l$.  Varying  $l \ll \lambda$ affects only the radiation resistance.

The Effective Area of a Receiving Antenna

How can we characterize antennas used for receiving, as in radio astronomy, rather than for transmitting? The receiving counterpart of transmitting power gain is the effective area or effective collecting area of an antenna.

Imagine an ideal antenna that collects all of the radiation falling on it from a distant point source and converts it to electrical power—a "rain gauge" for collecting photons.  The total spectral power that it collects will be the product of its geometric area $A$ and the incident spectral power per unit area, or flux density $S$.  By analogy, if any real antenna collects spectral power $P_\nu$, its effective area $A_{\rm e}$ is defined by
$$\bbox[border:3px blue solid,7pt]{ A_{\rm e} \equiv {P_\nu \over S_{\rm (matched)}}}\rlap{\quad \rm {(3A5)}}~,$$ where $S_{\rm (matched)}$ is the flux density in the "matched" polarization.

What does matched polarization mean? Any electromagnetic wave can be decomposed into two orthogonal polarized components. For example, the transverse electric field can be resolved into horizontal and vertical components, or horizontal and vertical linear polarizations. If the horizontal and vertical electric fields are equal in amplitude and $90^\circ$ out of phase, the radiation is circularly polarized. Any radio wave can also be decomposed into left- and right-circular polarizations. If the wave is essentially random (noise generated by blackbody radiation for example), the two orthogonal components will vary rapidly in intensity but have equal powers when averaged over long times. Such radiation is called unpolarized. Thus for an unpolarized source,
$$S_{\rm (matched)} = {S \over 2} ~.$$ Blackbody radiation is unpolarized.  Most radio astronomical sources are unpolarized or nearly so.

Any antenna with a single output collects only one of the two polarizations from an electromagnetic wave. For example, a linear dipole antenna collects radiation only from the linear polarization whose electric field is parallel to the antenna wires. Electric fields perpendicular to the dipole antenna do not produce currents in the antenna, so the linear dipole is completely insensitive to the linear polarization perpendicular to its wires.  A pair of crossed dipoles is needed to collect power from both orthogonal polarizations simultaneously.

Just as energy conservation implies that all lossless transmitting antennas have the same average power gain, all lossless receiving antennas have the same average collecting area.  This average collecting area can be calculated via another thermodynamic thought experiment.

A cavity in thermodynamic equilibrium at temperature $T$ containing a resistor $R$ is coupled to an antenna, also at temperature $T$, through a filter passing frequencies in the range $\nu$ to $\nu + d \nu$.

Imagine an antenna inside a cavity in full thermodynamic equilibrium at temperature $T$ connected through a transmission line to a matched resistor in a second cavity at the same temperature. Suppose further that the connection contains a filter that passes only a narrow range of frequencies between $\nu$ and $\nu + d \nu$. Since this entire system is in thermodynamic equilibrium, no net power can flow between the antenna and the resistor. Otherwise, one cavity would heat up and the other would cool down, in violation of the second law of thermodynamics. Thus the total spectral power from all directions collected by the antenna
$$P_\nu = A_{\rm e} S_{\rm (matched)} = A_{\rm e} {S \over 2} = \int_{4\pi} A_{\rm e}(\theta,\phi) {B_\nu \over 2} d\Omega$$ must equal the Nyquist spectral power $P_\nu = kT$ produced by the resistor. Using the Rayleigh-Jeans approximation
$$B_\nu = {2 k T \over \lambda^2}$$ gives $$P_\nu = {2 k T \over 2 \lambda^2} \int_{4 \pi} A_{\rm e} d \Omega = k T$$ $$\int_{4 \pi} A_{\rm e} d \Omega = \lambda^2$$
The average collecting area is defined by $$\langle A_{\rm e} \rangle \equiv {\int_{4 \pi} A_{ \rm e} d \Omega \over \int_{4 \pi} d \Omega }~.$$

The effective collecting area of a receiving antenna is independent of its radiation environment, so this result applies for any type of radiation, not just blackbody radiation.  Without using Maxwell's equations we have obtained the remarkable result true for all lossless antennas:
$$\bbox[border:3px blue solid,7pt]{\langle A_{\rm e} \rangle = {\lambda^2 \over 4 \pi}}\rlap{\quad \rm {(3A6)}}$$

• Any antenna, from a short dipole to the 100-m diameter Green Bank Telescope, has the same average collecting area $\langle A_{\rm e} \rangle$ that depends only on wavelength.

In the case of an isotropic antenna, the effective collecting area in any direction equals the average collecting area:
$$A_{\rm e}(\theta, \phi) = \langle A_{\rm e} \rangle = {\lambda^2 \over 4 \pi}$$ Consequently, a nondirectional antenna operating at a very short wavelength $\lambda$ will have a very small effective collecting area and poor sensitivity for reception. For this reason, most satellite broadcast services, GPS or satellite FM radio for example, transmit at relatively long wavelengths (10 to 20 cm). Likewise, practical radio telescopes can be constructed from arrays of dipoles at long wavelengths ($\lambda > 1$ m) but not at short wavelengths because the number of dipoles needed to produce useful collecting areas would be too large.

Reciprocity Theorems

Many antenna properties are the same for both transmitting and receiving. It is often easier to calculate the gain of a transmitting antenna than the collecting area of a receiving antenna, and it is often easier to measure the receiving power pattern than to measure the transmitting power pattern of a large radio telescope. Thus this receiving/transmitting "reciprocity" greatly simplifies antenna calculations and measurements. Reciprocity can be understood via Maxwell's equations or by thermodynamic arguments.

Burke & Smith (1997) state the electromagnetic case for reciprocity clearly: "An antenna can be treated either as a receiving device, gathering the incoming radiation field and conducting electrical signals to the output terminals, or as a transmitting system, launching electromagnetic waves outward. These two cases are equivalent because of time reversibility: the solutions of Maxwell's equations are valid when time is reversed."

The strong reciprocity theorem:

If a voltage is applied to the terminals of an antenna A and the current is measured at the terminals of another antenna B, then an equal current (in both amplitude and phase) will appear at the terminals of A if the same voltage is applied to B.

can be formally derived from Maxwell's equations [see a partial derivation in Rohlfs & Wilson Section 5.4] or by network analysis [see Kraus "Antennas", p. 252].

The strong reciprocity theorem implies that the transmitter voltages $V_{\rm A}$ and $V_{\rm B}$ are related to the receiver currents $I_{\rm A}$ and $I_{\rm B}$ by
$${V_{\rm A} \over I_{\rm B}} = {V_{\rm B} \over I_{\rm A}}$$
for any pair of antennas $A$ and $B$.

For most radio astronomical applications, we are not concerned with the detailed phase relationships of voltages and currents, and we can use a weak reciprocity theorem that relates the angular dependences of the transmitting power pattern and the receiving collecting area of any antenna:

The power pattern of an antenna is the same for transmitting and receiving.

That is:
$$\bbox[border:3px blue solid,7pt]{G(\theta, \phi) \propto A_{\rm e}(\theta, \phi)}\rlap{\quad \rm {(3A7)}}$$

The weak reciprocity theorem can be proven by another simple thermodynamic thought experiment: An antenna is connected to a matched load inside a cavity initially in equilibrium at temperature $T$. The antenna simultaneously receives power from the cavity walls and transmits power generated by the resistor. The total power transmitted in all directions must equal the total power received from all directions since no net power can be transferred between the antenna and the resistor; otherwise the resistor would not remain at temperature $T$. Moreover, in any direction, the power received and transmitted by the antenna must be the same, else the cavity wall in directions where the transmitted power was greater than the received power would rise in temperature and the cavity wall in directions of lower transmitted/received power ratio would cool, leading to a violation of the second law of thermodynamics.

The weak reciprocity theorem states that the transmitting and receiving power patterns of an antenna cannot differ as shown here, without violating the second law of thermodynamics.

The constant of proportionality relating $G$ and $A_{\rm e}$ can be derived from our earlier results for an isotropic antenna
$$\langle A_{\rm e} \rangle = {\lambda^2 \over 4 \pi} {\rm \qquad and \qquad} \langle G \rangle = 1$$

Thus energy conservation and the weak reciprocity theorem imply
$$\bbox[border:3px blue solid,7pt]{A_{\rm e}(\theta, \phi) = {\lambda^2 G(\theta, \phi) \over 4 \pi}}\rlap{\quad \rm {(3A8)}}$$ for any antenna. This extremely useful equation lets us compute the receiving power pattern from the transmitting power pattern and vice versa.

Example: We can use our calculation of the transmitting power pattern of a short dipole to calculate its effective collecting area when used as a receiving antenna:
$$A_{\rm e}(\theta,\phi) = {\lambda^2 G(\theta,\phi) \over 4 \pi} = {\lambda^2 \over 4 \pi} {3 \sin^2\theta \over 2}$$ $$A_{\rm e} = {3 \lambda^2 \sin^2 \theta \over 8 \pi}$$

Example: What is the power per unit bandwidth $P_\nu$ collected by a short dipole at $\nu = 10$ GHz broadside to ($\theta = 90^\circ$) the Sun, a thermal source whose flux density is $S \approx 1.2 \times 10^6$ Jy?

The broadside ($\sin\theta = 1$) collecting area of the short dipole is
$$A_{\rm e} = {3 \lambda^2 \over 8 \pi}$$ $$\lambda = {c \over \nu} = {3 \times 10^8 {\rm ~m~s}^{-1} \over 10^{10} {\rm ~Hz}} = 0.03{\rm ~m}$$ so $$A_{\rm e} = {3 \times (0.03{\rm ~m})^2 \over 8 \pi} = 1.07 \times 10^{-4}{\rm ~m}^2 \approx 1{\rm ~cm}^2~.$$
Clearly, dipoles or other nearly isotropic antennas have very small collecting areas at short wavelengths.  For an unpolarized source,
$$P_\nu = A_{\rm e} S_{\rm (matched)} = {A_{\rm e} S \over 2}$$ $$P_\nu = 1.07 \times 10^{-4} {\rm ~m}^2 \times {1.2 \times 10^6 {\rm ~Jy} \over 2} \times {10^{-26} {\rm ~W~m}^{-2}{\rm ~Hz}^{-1} \over 1{\rm ~Jy}}$$ $$P_\nu = 6.4 \times 10^{-25} {\rm ~W~Hz}^{-1}$$

Arrays of dipoles do make sense at long wavelengths.  For example, the Long Wavelength Array (LWA) for $\nu =$ 20 to 80 MHz ($\lambda \approx$ 4 to 15 m) will consist of 53 stations, each a 100 m $\times$ 100 m array of crossed dipoles.  The effective collecting area of the LWA will be up to $A_{\rm e} = 4 \times 10^6$ m$^2$ at $\lambda = 15$ m.

The proposed Long-Wavelength Array of crossed dipoles. Image credit

Eight test dipoles of the LWDA (Long-Wavelength Development Array) on the VLA site. Image credit

Antenna Temperature

A convenient practical unit for the power output per unit frequency from a receiving antenna is the antenna temperature $T_{\rm A}$. Antenna temperature has nothing to do with the physical temperature of the antenna as measured by a thermometer; it is only the temperature of a matched resistor whose thermally generated power per unit frequency equals that produced by the antenna. It is widely used because:

1. 1 K of antenna temperature is a conveniently small power. $T_{\rm A} = 1$ K corresponds to $P_\nu = k T_{\rm A} = 1.38 \times 10^{-23} {\rm ~J~K}^{-1} \times 1 {\rm ~K} = 1.38 \times 10^{-23} {\rm ~W~Hz}^{-1}$.
2. It can be calibrated by a direct comparison with hot and cold loads (another word for matched resistors) connected to the receiver input.
3. The units of receiver noise are also K, so comparing the signal in K with the receiver noise in K makes it easy to decide if a signal will be detectable.

$$\bbox[border:3px blue solid,7pt]{T_{\rm A} \equiv {P_\nu \over k} }\rlap{\quad \rm {(3A9)}}$$
An unpolarized point source of flux density $S$ increases the antenna temperature by
$$\bbox[border:3px blue solid,7pt]{T_{\rm A} = {P_\nu \over k} = {A_{\rm e} S \over 2 k} }\rlap{\quad \rm {(3A10)}}~$$
where $A_{\rm e}$ is the effective collecting area.  It is often convenient to express the point-source sensitivity of a radio telescope in units of "Kelvins per Jansky" rather than in units of area (m$^2$).  The effective area corresponding to a sensitivity of $1{\rm ~K~Jy}^{-1}$ is
$$A_{\rm e} = {2 k T \over S} = {2 \cdot 1.38065 \times 10^{-23} {\rm ~J~K}^{-1} \cdot 1{\rm ~K}\over 10^{-26} {\rm ~W~m}^{-2} {\rm ~Hz}^{-1}} = 2761 {\rm ~m}^2~.$$

Example: What is the increase in the antenna temperature of our short dipole produced by thermal emission from the Sun at $\nu = 10$ GHz?

$$P_\nu = 6.4 \times 10^{-25} {\rm ~W~Hz}^{-1}$$ so $$T_{\rm A} = {6.4 \times 10^{-25} {\rm ~W~Hz}^{-1} \over 1.38 \times 10^{-23} {\rm ~J~K}^{-1}} \approx 0.046 {\rm ~K}$$

Beam Solid Angle

The beam solid angle $\Omega_{\rm A}$ of a lossless antenna is defined as
$$\bbox[border:3px blue solid,7pt]{\Omega_{\rm A} \equiv \int_{4 \pi} P_{\rm n} (\theta, \phi) d \Omega}\rlap{\quad \rm {(3A11)}}$$
where $P_{\rm n} (\theta, \phi)$ is the power pattern normalized to unity maximum:
$$P_{\rm n} = {G(\theta, \phi) \over G_{\rm max}}~.$$ $$\Omega_{\rm A} = {1 \over G_{\rm max}} \int_{4 \pi} G d \Omega = {4 \pi \over G_{\rm max}}$$

The beam solid angle is a useful parameter for estimating the antenna temperature produced by a compact source covering solid angle $\Omega_{\rm s}$ and having uniform brightness temperature $T_{\rm B}$.  "Compact" means that the source is much smaller than the beam so that the variation of $P_{\rm n}$ is small across the source.  The power per unit bandwidth received from the source by the antenna pointing at it is
$$P_\nu = \int_{4 \pi} A_{\rm e}(\theta, \phi) {B_\nu \over 2} d \Omega$$ $$P_\nu \approx {\lambda^2 G_{\rm (max)} k T_{\rm B} \Omega_{\rm s} \over 4 \pi \lambda^2} = k T_{\rm B} {\Omega_{\rm s} \over \Omega_{\rm A}}~.$$ Thus the antenna temperature $T_{\rm A} = P_\nu / k$ is $$T_{\rm A} \approx T_{\rm B} {\Omega_{\rm s} \over \Omega_{\rm A}}~.$$ Stated in words, the antenna temperature equals the source brightness temperature multiplied by the fraction of the beam solid angle filled by the source.  A $T_{\rm B} = 10^4$ K source covering 1% of the beam solid angle will add 100 K to the antenna temperature.

Main Beam Solid Angle

The main beam of an antenna is defined as the region containing the principal response out to the first zero; responses outside this region are called sidelobes or, very far from the main beam, stray radiation.  The main beam solid angle $\Omega_{\rm MB}$ is defined as
$$\bbox[border:3px blue solid,7pt]{\Omega_{\rm MB} \equiv \int_{\rm MB} P_{\rm n} (\theta, \phi) d \Omega}\rlap{\quad \rm {(3A12)}}$$
The fraction of the total beam solid angle inside the main beam is called the main beam efficiency or, loosely, the beam efficiency.
$$\bbox[border:3px blue solid,7pt]{\eta_{\rm B} \equiv {\Omega_{\rm MB} \over \Omega_{\rm A}}}\rlap{\quad \rm {(3A13)}}$$


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# Finding Amplitude

## Homework Statement

An air-track glider attached to a spring oscillates between the 9.23 cm mark and the 50.23 cm mark on the track. The glider completes 27 oscillations in 44.7 s. angular frequency of the glider? (in rad/s)
3.7933
what is the amplitude of the glider?

x=Acos(wt+ro)

## The Attempt at a Solution

I considered that at x=9.23 the amplitude was zero and so is ro.
Hence, x=50.23 would equal just 50.23 cm=Acos(3.79(44.7)). But A turned out to be negative. What am I doing wrong? How do I solve this problem? Thank you in advance.

dx
Homework Helper
Gold Member
What is the center of oscillation here? The amplitude is the maximum distance from the center that the glider reaches.

Would it be half the distance the glider travels which is 20.5 cm?

dx
Homework Helper
Gold Member
Correct.

It still gives me a negative number if I do .205/((cos(3.79*44.7)).

dx
Homework Helper
Gold Member
Negative number for what?

For the amplitude A.

dx
Homework Helper
Gold Member
You just said the amplitude was 20.5 cm, which is correct.

OH! Ok I get it. Thank you very much.

format: plain


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# Calculate sums of logs in precision

I am encountering a situation where I cannot calculate exact sum of a seris of logorithms in calculating entropy. Suppose we have a series of numbers $p_i$ and we want to calculate $\sum_ilog(p_i)$, we should multiply $p_i$ together then take the log, or just take the log of each $p_i$ and calculate the summation over them?

The $p_i$ satisfies that $\sum_i p_i = 1$ and each $p_i$ is between 0 and 1.

Things are, we multiply them together, seems I will encounter the loss of precision and calculate the multiplication will accumulate the loss of precision. But when doing log apart, then each log seems to have loss of precisions.

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You would actually want to find $\sum_i p_i\log p_i$, know? In standard texts, they always find $\log$ of each $p_i$ and multiply by $p_i$ and sum it up. But I do not know which is better and why. – Ashok Oct 1 '12 at 12:44

Using standard computer floating point you will have errors at every step along the way, but they shouldn't be too bad. Your $p_i$'s probably cannot be represented exactly, so even $p_1+p_2$ has some error associated. In 64 bit floats, this amounts to a relative error of order $2^{-53}$, so even a lot of those rarely amount to much. It was worse in the 32 bit days. The log function will have errors, too, but if you use a decent library function it shouldn't be much worse. The risk when you multiply comes if you overflow the available range, but this is unlikely too.

The real problem usually comes when you subtract two nearly equal quantities. In your case the terms all have the same sign, so this is not a problem. If your terms cover several orders of magnitude, it is better to add the small ones first.

-
– leonbloy Oct 31 '14 at 10:22


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zbMATH — the first resource for mathematics

Müntz systems and orthogonal Müntz-Legendre polynomials. (English) Zbl 0799.41015
Summary: The Müntz-Legendre polynomials arise by orthogonalizing the Müntz system $$\{x^{\alpha_ 0}, x^{\alpha_ 1}, \dots\}$$ with respect to Lebesgue measure on [0,1]. In this paper, differential and integral recurrence formulae for the Müntz-Legendre polynomials are obtained. Interlacing and lexicographical properties of their zeros are studied, and the smallest and largest zeros are universally estimated via the zeros of Laguerre polynomials. The uniform convergence of the Christoffel functions is proved equivalent to the nondenseness of the Müntz space on [0,1], which implies that in this case the orthogonal Müntz-Legendre polynomials tends to 0 uniformly on closed subintervals of [0,1). Some inequalities for Müntz polynomials are also investigated, most notably, a sharp $$L^ 2$$ Markov inequality is proved.

MSC:
 41A17 Inequalities in approximation (Bernstein, Jackson, Nikol’skiĭ-type inequalities) 42C05 Orthogonal functions and polynomials, general theory of nontrigonometric harmonic analysis 30C15 Zeros of polynomials, rational functions, and other analytic functions of one complex variable (e.g., zeros of functions with bounded Dirichlet integral) 39A10 Additive difference equations
Full Text:


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### Fitting In

The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ

### Look Before You Leap

Can you spot a cunning way to work out the missing length?

### Triangle Midpoints

You are only given the three midpoints of the sides of a triangle. How can you construct the original triangle?

# Points in Pairs

##### Age 14 to 16 Challenge Level:

Two points, one inside a circle and the other outside, are related in the following way :

A line starting at the centre of the circle and passing through the first point ($P$) goes on to pass through the second point ($P'$)

Positions along the line are such that the ratio of $OP$ to the radius of the circle matches the ratio of the radius of the circle to $OP'$

For example if $OP$ happened to be $2/3$ of the radius then $OP'$ would be $3/2$ of the radius.

You can use the interactivity below to help you explore how the positions of a pair of points relate to each other.

Full Screen Version
If you can see this message Flash may not be working in your browser
Please see http://nrich.maths.org/techhelp/#flash to enable it.

Once you have a feel for that, leave the interactivity and go on to the main problem which is best done by drawing sketches on paper.

In the diagram above, $P$ and $P'$ are a connected pair, and $Q$ and $Q'$ the same, both pairs behaving like the points in the interactivity.

Now for the problem : The radius of the circle is 10 units, $OP$ is 8 units and $OQ$ is 6 units.

If the distance $PQ$ is 5 units what is the distance $P'Q'$?


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## Algebra 1: Common Core (15th Edition)

B: y=$\frac{1}{2}$x+2
The given line has a slope of $\frac{1}{2}$. If two lines are parallel they have the same slope so the other slope is $\frac{1}{2}$.Put the equation into the point-slope form using the point (-2,1) and the slope of 1/2: y-y1=m(x-x1) y-1=$\frac{1}{2}$(x-(-2))


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# AP 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

These AP 10th Class Physical Science Chapter Wise Important Questions 10th Lesson Chemical Bonding will help students prepare well for the exams.

## AP State Syllabus 10th Class Chemistry 10th Lesson Important Questions and Answers Chemical Bonding

### 10th Class Chemistry 10th Lesson Chemical Bonding 1 Mark Important Questions and Answers

Question 1.
What is a chemical bond? (AP March 2015)
An attractive force between two atoms in a molecule is called a chemical bond.

Question 2.
Write the names of any two compounds which have ionic bond. (TS June 2016)
1) NaCl
2) MgCl2

Question 3.
Draw the structural diagram of Ammonia molecule as per the valence-shell electron pair repulsion theory. (TS March 2016)

Question 4.
Show the formation of HC/ molecule with Lewis dot structures using the information given below. (TS March 2017)

Question 5.
Explain, why bonding angle (HOH) in water is 104° 31′ instead of 109° 28′? (TS June 2018)
Due to the repulsion between the lone pair of electrons and bond pair of electrons in water molecule the bond angle will be 104° 31′ rather than 109° 28′.

Question 6.
Imagine and write what type of ion can be formed generally by an atom of element with low ionisation energy, low electron affinity with high atomic size? (AP March 2019)
Cation or positivity charged ion can be formed generally by an atom of element with low ionisation energy, low electron affinity with high atomic size.

Question 7.
Draw the structure of Methane molecule and mention bond angle. (TS March 2019)
Structure of methane CH4. Tetrahedral structure with bond angle 109°28′.

Bond Angle is 109°28′

Question 8.
Why do elements form chemical bonds? (TS SCERT: 2019-20)

• Elements are unstable if they contain less than eight electrons (octet) in their valency shell.
• Hence, they form chemical bond with other elements to get octet in their valency shell.

Question 9.
How is covalent bond formed?
A covalent bond is formed by the sharing of electrons between two atoms.
Ex : Bonding in H2 molecule, O2 molecule, N2 molecule, etc.

Question 10.
What is ‘Octet rule’?
Octet rule :
Presence of 8 electrons in the outermost shell of an atom or a molecule is called ‘octet rule’.

Question 11.
What is ‘Bond length’?
Bond length :
It is the inner-nucleus distance between the two atoms in a molecule. It is measured in Angstrom, 1 Å = 10-8 cm.

Question 12.
How is a cation formed?
A cation is formed when an atom loses electrons.

Question 13.
How is an anion formed?
An anion is formed when an atom gains electrons.

Question 14.
Which type of compounds are more soluble in polar solvents?
Ionic compounds are more soluble in polar solvents.

Question 15.
Which compounds exhibit high melting and boiling points?
Ionic compounds exhibit high melting and boiling points.

Question 16.
What is electronic configuration?
A systematic arrangement of electrons in the atomic orbits is called electronic configuration.

Question 17.
Why are molecules more stable than atoms?
Molecules have lower energy than that of the combined atoms. Molecules are more stable than atoms since chemical species with lower energy are more stable.

Question 18.
When two oppositely charged ions are engaged in a bond, it is known as ionic linkage.

Question 19.
On which factors do anions depend?

1. Atomic size
2. Ionisation potential
3. Electron affinity
4. Electronegativity

Question 20.
How do you know the valence of a metal?
The number of electrons lost from a metal atom is the valence of its element which is equal to its group number.
Ex : Na and Mg have valence 1 and 2 respectively.

Question 21.
How do you know the valence of a non-metal?
The number of electrons gained by a non-metal element for its atom is its valency, which is equal to 8 – its group number. Ex : The valency of chlorine is (8 – 7) = 1.

Question 22.
What is Tonic bond’?
The electrostatic attractive force that keeps cation and anion together to form a new electrically neutral compounds is called ionic bond.

Question 23.
Why do atoms combine and form molecules?
The energy of molecule is less than the total energy of constituent atoms. Therefore atoms combine and go to a stable state of lower energy.

Question 24.
What is ‘orbital concept of bond formation’?
Atoms with half-filled or vacant orbitals try to get paired electrons in those orbitals by bond formation, i.e. by losing, gaining or sharing of electrons.

Question 25.
Name the bonds present in the molecules
i) BaCl2
ii) C2H4.
i) In BaCl2 – Ionic.
ii) In C2H4 – Covalent (double bond H2C = CH2).

Question 26.
Why are ionic compounds good electrolytes?

• Electrolytes produce ions in solution, which carry current.
• Ionic compounds in the fused state and aqueous solutions contain ions moving freely. Hence they conduct electricity.

Question 27.
When is ionic bond formed between atoms?
Ionic bond is readily formed between atoms of elements with a low ionisation energy and atoms of elements with high electronegativity.

Question 28.
What is crystal lattice’?
In a crystal of an ionic compound, each ion is surrounded by oppositely charged ions. The ions arrange themselves at an optimum distance with regular periodicity in a well-defined three-dimensional network called crystal lattice.

Question 29.
What is Lattice energy’?
The energy released when gaseous positive and negative ions are brought together from infinity to form one mole ionic crystals is called lattice energy.

Question 30.
How is a σ (sigma) bond formed?
By the axial or head on overlap of pure orbitals or hybrid orbitals of two atoms.

Question 31.
How is a π (pi) bond formed?
By the lateral or parallel overlap of only pure atomic orbitals after the a bond formation.

Question 32.
Which type of atoms easily enter ionic bonding?
An atom with low ionization potential and another with high electron affinity.

Question 33.
What is a polar bond?
A covalent bond in a heteroatomic molecule.
Eg : HCl.

Question 34.
What is meant by inter-nuclear axis?
The hypothetical line joining the centre of nuclei of two atoms in a molecule.

Question 35.
What are multiple bonds?
Double and triple bonds are multiple bonds.

Question 36.
How many a and π bonds are in O2 molecule?
One σ and one π bond.

Question 37.
What are Lewis structures?
The symbol of the atom gives the core (or kernel) of the atom, in which valence electrons are shown as dots (•), circles (O) or crosses (x). Thus in Na, Na represents the core of sodium atom and the cross (x) represents the valence electron of sodium.

Question 38.
What are the structures of sodium chloride and calcium chloride crystals?
Sodium chloride has face centered cubic structure. Calcium chloride has body centered cubic structure.

Question 39.
NaCl dissolves in water but not in benzene. Explain.
NaCl dissolves in water because of hydration. Water being a polar molecule has positive and negative ends which hydrate Na+ and Cl ions. Benzene being non-polar cannot solvate the ions of NaCl.

Question 40.
What are the bond angles in H2O and NH3 molecule?
Bond angle in H2O molecule is 104° 30′.
Bond angle in NH3 molecule is 107°.

Question 41.
What are the forces present in an ionic bond?
Electrostatic forces of attraction are present in ionic bonds.

Question 42.
Which forces are weaker forces and where are they operative?

1. Van der Waal’s forces are very weak forces.
2. They are operative between non-polar molecules.

Question 43.
Which compounds exhibit low melting and boiling points?
Covalent compounds exhibit low melting and boiling points when compared to ionic compounds.

Question 44.
How are cations formed?
The metal atoms lose electrons to form positively charged ions or cations.

Question 45.
Why is a molecule of hydrogen more stable than the uncombined atoms?
When a molecule of hydrogen is formed from the atoms, energy is released (104 Kcal/mol). Thus the molecules possessing lower energy are more stable than the atoms.

Question 46.
How many sigma and pi bonds are present in acetylene molecule between carbon atoms?
$$\mathrm{HC} \equiv \mathrm{CH}$$, one sigma and two pi bonds are present.

Question 49.
In case of ionic substances, a more appropriate term is formula weight, rather than ‘molecular weight’. Why?
Molecules are not present in ionic substances. Only ions are present. The formula of an ionic substance represents the simpler ratio of ions in one mole of crystal. Hence a more appropriate term is ‘formula weight’.

Question 50.
Write a short note on bond angles.
It is the average angle between two adjacent atoms bonded to the central atom in a molecule. Molecules with larger bond angles are more stable than those with smaller angles.

Question 51.
Sulphur dioxide is a diamagnetic molecule. Explain.
Sulphur dioxide (SO2) is a diamagnetic molecule because it has all electrons paired (no free electrons).

Question 52.
What is an ion?
An ion is an electrically charged atom (or group of atoms).

Question 53.
What is a Coordination number?
The number of ions of opposite charge that surround a given ion of given charge is known as coordination number of that given ion. –

Question 54.
What factors affect or which factors influence the formation cation or anion?
Factors affecting formation of anion or cation are :

1. Atomic size,
2. Ionization potential,
3. Electron affinity,
4. Electronegativity.

Question 55.
Two elements X and Y have the following configurations.
X = 1s² 2s² 2p6 3s² 3p6 4s²
Y = 1s² 2s² 2p6 3s² 3p5
What is the formula of the compound?
The electronic configuration of X is 1s² 2s² 2p6 3s² 3p6 4s².
So its valency is 2.
The electronic configuration of Y is 1s² 2s² 2p6 3s² 3p5.
So its valency is 1.
∴ The formula of the compound is XY2.

Question 56.
The electronegativities of two elements are 1 and 3. What type of bond is formed between the two elements? Why?
The electronegativity difference of elements = 3 – 1 = 2.
So the electronegativity difference is more than 1.9. Therefore the bond formed between the elements is ionic in nature.

Question 57.
What do you mean by Doublet configuration?
If two electrons are present in valence shell, then it is called doublet configuration.
Eg.: Helium exhibits doublet configuration.

Question 58.
Write Lewis symbol of potassium and calcium.
Potassium – $$\dot{\mathrm{K}}$$ and Calcium – •Ca•

Question 59.
Which of the following is / are true?
1) Ionic compounds exist as solids.
2) Ionic compounds have high melting and boiling points.
3) Ionic compounds conduct electricity in solid state.
4) Ionic compounds have low melting and boiling points.
Ionic compounds generally exists as solids and they have very high melting and boiling points and they conduct electricity in aqueous or molten state.
So first and second statements are true.

Question 60.
How many a and n bonds are there in acetylene molecule?
The structure of acetylene is given by
$$\mathrm{H}-\mathrm{C} \equiv-\mathrm{H}$$
So they are 3a and 2n bonds.

Question 61.
What is the valency of carbon in CO2?
The valency of oxygen is 2. So the valency of carbon = 2 × 2 = 4.

Question 62.
Which of the following does not get Helium configuration during formation of chemical bonding?
1) Hydrogen,
2) Lithium,
3) Beryllium,
4) Oxygen.
Oxygen attains Neon configuration when it undergoes chemical bondings and the rest will attain Helium configuration.

Question 63.
Write the valencies of following elements.
a) Be
b) N and what is the compound formed when these two reacted?
a) The electronic configuration of Be is 1s² 2s². So its valency is 2.
b) The electronic configuration of Nitrogen is 1s² 2s² 2p³. So its valency is 3.
∴ Formula of the compound is Be3N2.

Question 64.
Which of the following solutions are conductors of electricity?
a) Sugar solution,
b) Alcohol solution,
c) Glucose solution,
d) Salt solution.
Salt solution is good conductor of electricity because common salt (NaCl) is an ionic compound and the rest are covalent compounds.

Question 65.
Why are noble gases (inert gases) stable?

1. Noble gases are stable because their outermost orbit contains 8 or 2 electrons.
2. Noble gases exist an individual atoms.

### 10th Class Chemistry 10th Lesson Chemical Bonding 2 Marks Important Questions and Answers

Question 1.
Represent each of the following molecules using Lewis notation. (TS March 2015)
(i) Calcium and Chlorine to form Calcium chloride.
(ii) Formation of Oxygen molecule from Oxygen atoms.

Question 2.
Between a neutral atom and its cations which has bigger size? Why? (TS June 2016)

• A neutral atom has bigger in size than its cation.
• A cation has more protons and nucleus of the cation attracts the electrons in the outermost orbital more.
• Hence, the radius of the anion decreases. It means size of the anion decreases.
• So, a neutral atom is bigger than its cation.

Question 3.
Draw the diagram to show the formation of Oxygen molecule by Valence bond theory. (TS March 2017)

Question 4.
Explain Ionic bond with suitable example. (TS June 2018)
1) Sodium (Na) looses one electron and forms Sodium ion (Na+).
Na → Na+ + 1e

2) Chlorine (Cl) gains one electron and forms chloride ion (Cl).
Cl + 1e → Cl

3) Positive sodium ion (Na+) and negative chloride (Ct) ion come together due to electrostatic forces, participates in ionic bond and form NaCl.
Na+ + Cl → NaCl

Question 5.
Give two examples of each to ionic and covalent compounds. (AP SCERT: 2019-20)
1) Ionic compounds eg : NaCl, MgCl2
2) Covalent compounds eg : Cl2, HCl

Question 6.
Distinguish between a sigma and a pi bond.

 Sigma bond Pi bond 1. It is formed by the end-end on overlap of orbitals. 1. It is formed by the lateral overlap of orbitals. 2. It has independent existence. 2. It has no independent existence. 3. It is a strong bond. Because axial overlap is more. 3. It is a weak bond. Because lateral overlap is less. 4. There can be only one a bond. 4. There can be one or two 7t bonds between two atoms. 5. All orbitals form ‘o’ bond. 5. Only p, d orbitals form a bond.

Question 7.

• Hydrogen bond is formed between molecules in which hydrogen atom is attached to an atom of an element with large electronegativity and very small size (F, O, N). Because in hydrogen bond the molecules associate themselves and hence possess higher B.P’s and M.P’s.
• The hydrogen bond formed between two molecules is called inter-molecular hydrogen bond.
• The hydrogen bond formed between different groups of the same molecule is called intra-molecular hydrogen bond.

Question 8.
Bring out the difference between ionic and covalent bonds.

 Ionic bond Covalent bond 1. It is formed by transference of electrons from one atom to the other. 1. It is formed by the sharing of electron pairs by two atoms. 2. Electrostatic. 2. Not electrostatic, but rigid. 3. Ionic substances are formed by ionic bonds. 3. Molecules are formed by covalent bonds. 4. Non-directional. 4. Directional.

Question 9.
What is hybridisation?
In the formation of molecules, the atomic orbitals of the atoms may hybridise.

1. It is the process of mixing up of atomic orbitals of an atom to form identical hybrid orbitals.
2. This takes place only during the formation of bond.
3. There should not be much difference in the energies of the orbitals that hybridise.
4. The number of hybrid orbitals formed is equal to the number of hybridising atomic orbitals.
5. Hybrid orbitals form a bonds only not n bonds.

Question 10.
What is ionisation? Give one example.
1) The process of removal of electron (s) from an atom or molecule is termed as ionisation.
Eg : Na → Na+ + e

2) Dissociation of an ionic solid into constituent ions upon its dissolution in a suitable solvent, is also called ionisation.
Eg : NaCl(aq) → Na+(aq) + Cl(aq)

Question 11.
Write a short note on octet rule with example.
Atoms try to get 8 electrons in their outermost shell (inert gas structure) by combining with other atoms. In order to get 8 outer electrons, atoms may lose to, gain from or share electrons with other atoms.

Question 12.
What are the important characteristic features of hybridisation?

• Orbitals on a single atom only would undergo hybridization.
• The orbitals involved in hybridisation should not differ largely in their energies.
• The number of hybrid orbitals formed is equal to the number of hybridising orbitals.
• The hybrid orbitals form stronger directional bonds than the pure s, p, d atomic orbitals.
• It is the orbitals that undergo hybridisation and not the electrons.
• Concept of hybridisation is useful in explaining the shape of molecules.

Question 13.
Why is a large amount of energy needed to remove an electron from a neutral gaseous neon atom than the energy needed to remove an electron from gaseous sodium atom?
Na (g) +118.4 K.cal → Na+ + e
Ne (g) + 497.0 K.cal → Ne+ + e

Orbitals which are fully-filled are very stable, so need large amount of energy to remove an electron from them.

Question 14.
What are the drawbacks of electronic theory of valence?
1) When covalent bond formed between any two atoms, irrespective of the nature of the atoms, the bond lengths and bond energies’ are expected to be same. But practically it was observed that bond lengths and bond energies are not same when the atoms that form the bond are different.

2) This theory fails to explain the shapes of molecules.

Question 15.
What is the structure of NaCl and write its coordination number of its constituent ions?
The structure of NaCl is face centred cubic lattice.
Its coordination number is 6 for Na+ and Cl.

Question 16.
Why is there absorption of energy in certain chemical reactions and release of energy in other reactions?
If bond dissociation energy of reactants is more than bond energy of products, then energy is absorbed in the chemical reaction.

If bond dissociation energy of reactants is less than bond energy of products, then energy is released in the chemical reaction.

Question 17.
Why do ionic compounds dissolve in polar solution and covalent compounds dissolve in non-polar solution?
Ionic compounds are polar in nature so they are soluble in polar solvents whereas covalent compounds are non-polar in nature so they are soluble in non-polar solvents.

Question 18.
Identify the following whether they are either oxidation or reduction reactions.

i) Oxygen atom is gaining electrons. So it is reduction reaction.
ii) Potassium atom is losing electron. So it is oxidation reaction.
iii) Ferric ion is gaining electron. So it is reduction reaction.

Question 19.
Elements A, B, and C have atomic numbers 9, 20, and 10 respectively.
a) State which one is (i) a metal, (ii) a non-metal, (iii) chemically inert.
b) Write down the formula of the compound formed by two of the above elements.
a) i) Element with atomic number 20 is a metal because its electronic configuration is 2, 8, 8, 2. It belongs to 2nd group and the element is calcium.
ii) Element with atomic number 9 is a non-metal because its electronic configuration is 2, 7. So it belongs to 17th group and it is Fluorine.
iii) Element with atomic number 10 is inert gas because its electronic configuration is 2, 8 and the element is Neon.

b) The formula of compound formed between elements Calcium and Fluorine is CaF2.

Question 20.
In the formation of the compound XY2, an atom X gives one electron to each Y atom. What is the nature of bond in XY2? Give two properties of XY2.
Two electrons are transferred from X to Y. X forms positive ion and Y forms a negative ion. So the bond formed is ionic in nature.
Properties of ionic compounds :
a) They are hard crystals.
b) They have high boiling point and melting point.
c) They are soluble in water.

Question 21.
Element X is a metal with valency 2. Element Y is a non-metal with valency 3.
i) Write equations to show X and Y form ions.
X is a metal with valency 2.

ii) If Y is a di-atomic gas, write the equation for the direct combination of X and Y to form a compound.
4X + 3Y2 → 2X2Y3

Question 22.
What are the differences between sodium atom and sodium ion?

• Sodium atom is neutral whereas sodium ion has unipositive charge.
• The size of sodium ion is smaller than sodium atom.
• The properties of sodium ion is different when compared with sodium.

Question 23.
What is the percent of p-character in sp, sp² and sp³ hybrid orbitals?
In sp the p-character is 50%.
In sp² the p-character is 66.66%.
In sp³ the p-character is 75%.

Question 24.
What is the percent of s-character in sp, sp² and sp³ hybrid orbitals?
In sp hybrid orbitals s-character is 50%.
In sp² hybrid orbitals s-character is 33.33%.
In sp³ hybrid orbitals s-character is 25%.

Question 25.
A chemical compound has following Lewis notation.

a) How many valence electrons does element Y have?
b) What is the valency of element Y?
c) What is the valency of element X?
d) How many covalent bonds are there in the molecule?
e) Suggest a name for the element X and Y.
a) The valence electrons in Y are 5.
b) The valency of Y is 3.
c) The valency of X is 1.
d) There are three covalent bonds.
e) The element Y is Nitrogen and X is Hydrogen.

Question 26.
The electronic configurations of three elements.
X is 2, 6; Y is 2, 8, 7; Z is 2, 8,1.
In each ease given below state whether the bonding is ionic or covalent and give the formula of molecules of the compound formed,
a) Between two atoms of X.
As we move from left to right in a period non-metallic character increases. So Z is a metal and remaining two are non-metals.
Therefore between two atoms of X, there would be covalent bond. The formula of compound is X2.

b) Between the atom X and atom Z.
X is a non-metal and Z is a metal. So between these two atoms, there would be an ionic bond.
The valency of X is 2 and the valency of Z is 1.
So the formula of compound is Z2X.

c) Between the atom Y and atom Z.
Y is also non-metal and we know Z is metal.
The valency of Y is 1.
So the compound formed is ZY.

Question 27.
Electronic configuration of X is 2, 8, 1 and electronic configuration of Y is 2, 8, 7. Explain what type of bond is formed between them.
The element with electronic configuration 2, 8,1 is sodium which is a metal. Similarly the element with electronic configuration 2, 8, 7 is chlorine which is a non-metal. The electronegativity difference between these two elements is more than 1.9 so they form ionic bond.

Question 28.
Complete the following statements.
i) The bond which is formed by transfer of an electron from one atom A to another atom B is called ……………
Ionic bond

ii) The atoms A and B become ions. Thus A – 2e = …………….. and B + e = ………….

iii) The formula of the compound is
The valency of A is 2 and valency of B is 1. So the compound formed is AB2.

Question 29.
i) What is the valency of Nitrogen in Ammonia?
The formula of Ammonia is NH3. We know valency of hydrogen is 1. So valency of Nitrogen is 3.

ii) What is the valency of Oxygen in Water?
The formula of water is H2O. The valency of hydrogen is 1. So the valency of oxygen is 2.

iii) What is the valency of Carbon in Carbon tetra chloride?
The formula of carbon tetra chloride is CCl4. The valency of chlorine is 1. So the valency of carbon is 4.

### 10th Class Chemistry 10th Lesson Chemical Bonding 4 Marks Important Questions and Answers

Question 1.
What is Hybridisation? Explain the formation of BeF, molecule using hybridisation. (AP June 2017)
Hybridisation :
Hybridisation is a phenomenon of intermixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

Formation of BeF2 molecule :

1. 4Be has electronic configuration 1s² 2s².
2. In excited state electronic configuration of 4Be is 1s² 2s¹ 2px¹
3. 4Be allows its 2s orbitals and 2px orbitals which contain unpaired electrons to intermix and redistribute to two identical orbitals.
4. The hybridisation is called ‘sp’ hybridisation and form two ‘sp’ hybrid orbitals. Each orbital contains one electron.
5. 9F has electronic configuration 1s² 2s² 2p5 (or) 1s² 2s² 2px² 2py² 2pz¹
6. F contains one unpaired electron.
7. Two Fluorine atoms come closure to Be and form two covalent bonds.
σsp – p and σsp – p
8. Now the BeF2 has (F $$\widehat{\mathrm{Be}}$$ F) angle of 180°.

Question 2.
The arrangement of electrons in different shells of atoms of 18th Group elements is given in the table.

Answer the following : (AP March 2017)
(i) What is the general electronic configuration of the above elements except He?
(ii) What is the valency of Argon?
(iii) Write Lewis dot structure of Neon.
(iv) Why the above elements do not take part in bond formation?
(i) ns² np6
(ii) Zero
(iii)

(iv) They are stable as they have 8 electrons (except Helium) in their outer most orbit.

Question 3.
“Nitrogen and Hydrogen react to form a molecule of Ammonia (NH3). Carbon and Hydrogen react to form a molecule of Methane (CH4).”
For each reaction :
a) What is the valency of each of the atom involved in the reaction?
b) Draw the dot structure of the products that are formed. (AP June 2018)
a) In ammonia : Nitrogen valency – 3, Hydrogen valency – 1
In methane : Carbon valency – 4, Hydrogen valency – 1

b)

Question 4.
Explain the formation of Boron tri-fluoride molecule by Hybridization. (TS June 2018)
1) Electron configuration of Boran at ground state.

2) Electron configuration of Boran at excited state is

3) As it forms three identical B-F bonds in BF3.
4) In excited state of Boran, one ‘s’ orbital and two ‘p’ orbitals undergoes hybridization and forms three sp² orbitals.
1s + 2p → 3sp²
5) Three sp² orbitals make end – on – end overlap with three ‘p’ orbitals of fluorine atoms.
6) They form three sp² – p sigma bonds.
7) In this way Boran tri-fluoride molecule is formed.
8)

Question 5.
Explain the formation of BF3 molecule with the help of Valence Bond theory? (TS March 2018)
Formation of BF3 molecule :

1. 5B has electronic configuration 1s² 2s² 2px.
2. The excited electronic configuration of 5B is 1s² 2s¹ 2px¹ 2py¹.
3. As it forms three identical B-F bonds in BF3
4. It is suggested that excited ‘B’ atom undergoes hybridisation.
5. There is an intermixing of 2s, 2px, 2py orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals.
6. For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one electron.
7. Now three fluorine atoms overlap their 2pz orbitals containing unpaired electrons (F9 1s² 2s² 2px² 2py² 2pz²) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three ssp²-p bonds.

Question 6.
Explain the formation of N2 molecule using Valence Bond theory. (TS June 2019)
Formation of N, molecule (Valence Bond theory) :

1. 7N has electronic configuration 1s² 2s² 2px¹ 2py¹ 2pz.
2. Suppose that px orbital of one ‘N’ atom overlaps the ‘px‘ orbital of the other ‘N’ atom giving σ px – px bond along the inter – nuclear axis.
3. The py and pz orbitals of one ‘N’ atom overlap the py and pz orbital of other ‘N’ atom laterally, respectively perpendicular to inter – nuclear axis giving π py – py and π pz-pz bonds.
4. Therefore, N2 molecule has a triple bond between two nitrogen atoms.

Question 7.
If the electronic configurations of atoms A and B are 1s², 2s², 2p6, 3s², 3p¹ and 1s², 2s², 2p4 respectively, then
a) which atom forms negative ion?
Given electronic configuration of atom A is 1s² 2s² 2p6 3s² 3p¹, i.e. Aluminium and B is 1s² 2s² 2p4, i.e. Oxygen.

The atom ‘B’ tends to form negative ion by gaining two electrons in order to get nearest inert gas. Neon’s configuration is 1s² 2s² 2p6. So its valency is 2.

b) which atom forms positive ion?
The atom ‘A’ tends to form positive ion by losing three electrons in order to get nearest inert gas. Neon’s configuration is 1s² 2s² 2p6.

c) what is the valency of atom A?
Valency of atom ‘A’ is ‘3’.

d) what is the molecular formula of the compound formed by atoms A and B ?
According to Criss-Cross method, the molecular formula of the compound formed by atoms both A and B is A2B3, i.e. Al2O3.

Question 8.
Writs the salient features of VSEPRT.
(OR)
Explain VSEPR theory.
The full form of VSEPRT is Valence Shell Electron Pair Repulsion Theory.

1. VSEPRT considers electrons in the valence shells which are in covalent bonds and in lone pairs as charge clouds that repel one another and stay as far apart as possible which will give specific shapes to molecules.
2. The presence of lone pair in the central atom causes slight distortion of bond angles from expected regular shape.
3. If the angle between lone pair and bond pair increases at the central atom due to more repulsion, the actual bond angles between atoms must be decreased.
4. If two bond pairs are present without any lone pair, the shape of the molecule is linear with bond angle 180°.
5. If three bond pairs are present without any lone pair, the shape of the molecule is trigonal planar with bond angle 120°.
6. If there are four bond pairs in the valency shell of central atom without lone pair, the shape is tetrahedron and the bond angle is 109°28′.
7. If there are three bond pairs and one lone pair, the shape of the molecule is pyramidal.
8. If there are two bond pairs and two lone pairs, the shape is V.

Question 9.
Do you think that a pair of Na+Cl as units would be present in the solid crystal? Explain.

• No, the electrostatic forces are non-directional.
• Therefore, it is not possible for one Na+ to be attracted by one Cl and vice-versa.
• Depending upon the size and charge of particular ion, number of oppositely charged ions get attracted by it, but, in a definite number.
• In sodium chloride crystal each Na+ is surrounded by 6 Cl and each Cl by six Na+ ions.
• Ionic compounds in the crystalline state consists of orderly arranged cations and anions held together by electrostatic forces of attractions in three dimensions.

Question 10.
Explain the valence bond theory.
Valence bond theory :

1. A covalent bond between two atoms is formed when the two atoms approach each other closely and one atom overlaps its valence orbital containing unpaired electron with the valence orbital of other atom that contains the unpaired electron of opposite spin.
2. The greater the overlapping of the orbitals, the stronger will be bond.
3. Valence bond theory gives a directional character to the bond when other than ‘s’ orbitals are involved.
4. Electron pair in the overlapping orbitals is shared bv both the atoms involved in the overlapping.
5. If orbitals overlap along inter-nucleus axis, they form a strong bond called sigma (σ) bond.
6. If the orbitals overlap laterally, they form a weak bond called pi (π) bond.

Question 11.
a) Name the charged particles which attract one another to form ionic or electro-valent compounds.
The charged particles are ions which are formed due to transfer of electrons from atom of one element to atom of other element.

b) In the formation of ionic compounds, electrons are transferred from one element to another. How are electrons involved in the formation of covalent compounds?
In the formation of covalent bond electrons are mutually shared between atoms of elements.

c) The electronic configuration of Nitrogen is 2, 5. How many electrons in the outer- shell of a nitrogen atom are not involved in the formation of Nitrogen molecule?
In the formation of Nitrogen molecules three pairs of electrons are mutually shared by the two nitrogen atoms. So each Nitrogen has two electrons which do not take part in bonding. Those two electrons are called lone pair.

d) In the formation of magnesium chloride, name the substance that is oxidized and the substance that is reduced.
Two electrons are transferred from Magnesium atom to Chlorine atom in the formation of Magnesium chloride.

So Magnesium is oxidised (loss of electrons is oxidation) and Chlorine is reduced (gain of electrons is reduction).

Question 12.
The symbols of two elements with their atomic numbers are given below. 16A and 20B
i) Which element will form a cation?
The atomic number of B is 20. So its electronic configuration is 1s² 2s² 2p6 3s² 3p6 4s² or 2, 8, 8, 2. So it easily loses two electrons to form cation thereby attain inert gas stability.

ii) Which element will form an anion?
The atomic number of A is 16. So its electronic configuration is 1s² 2s² 2p6 3s² 3p4 or 2, 8, 6. So it easily gains two electrons to form anion. Therefore it attains inert gas stability.

iii) Show how each element forms an ion.

iv) Which element will be oxidised in forming the ion?
B is losing electrons. So it is oxidised. .

v) When the elements react, what will be the nature of the bond formed?
When the elements react they form ionic bond because B is metal and A is non-metal.

Question 13.
Give one example for each of the following.
i) A polar covalent bond formed from two dissimilar atoms one of which is oxygen.
In the formation water molecule, polar covalent bond is formed between two dissimilar atoms hydrogen and oxygen.

ii) A non-polar covalent compound formed from two similar atoms.

The examples for a non-polar covalent compound formed from two similar atoms are Fluorine (F2), Chlorine (Cl2).

iii) A solid non-polar covalent compound.
The example for solid non-polar covalent compound is Iodine (I2).

iv) A polar covalent compound containing three shared pairs of electrons.

The polar covalent compound containing three shared pairs of electrons is Ammonia.

Question 14.
The list of some substances is given below.
HCl, H2O, Cl2, NaBr, CH4, NH3, N2, O2, CaO, HF, I2 and Br2.
From the list above choose the substance / substances.

i) Which have only ionic bond in the molecules?
NaBr, CaO have only ionic bond in the molecules,

ii) Which has a triple covalent bond in its molecule?
N2 has a triple covalent bond in its molecule ($$\mathbf{N} \equiv \mathbf{N}$$).

iii) Which are solids?
Generally ionic compounds are solids. So NaBr, CaO are solids. I2 being a covalent compound still it is solid.

iv) Which has a double covalent bond in its molecule?
O2 has a double covalent bond in its molecule (O = O).

v) Which is non-polar covalent liquid?
The non-polar covalent liquid is Br2.

Question 15.
There are elements with atomic numbers 4, 14, 8, 15 and 19. From this information answer the following questions.
a) A solid non-metal of valency 3.
The element with atomic number 15 is phosphorus. Its electronic configuration is 2, 8, 5. So its valency is 3 and it is a solid non-metal.

b) A gas of valency 2.
The element with atomic number 8 is oxygen. Its electronic configuration is 2, 6. So its valency is 2 and it is a gas.

c) A metal of valency 1.
The element with atomic number 19 is potassium. Its electronic configuration is 2, 8, 8, 1. So its valency is 1 and it is a metal.

d) A non metal of valency 4.
The element with atomic number 14 is silicon. Its electronic configuration is 2, 8, 4. Its valency is 4 and it is a non-metal.

Question 16.
Explain the following.
a) Ionic compounds conduct electricity.
Ionic compounds can be electrolysed to give their constituent ions. The ions obtained by dissociation move freely in solution and hence conduct electricity. When ionic compound is dissolved in water or melted, it becomes a good conductor of electricity.

b) Ionic compounds have high melting and boiling points while covalent compounds have low melting and boiling points.
The forces existing between ionic compounds are electrostatic in nature. They are strong forces. So in order to break these forces lot of energy is required. Therefore ionic compounds have higher boiling and melting points.

c) Ionic compounds dissolve in water whereas covalent compounds do not.
Ionic compounds are polar in nature. So they are soluble in polar solvent like water whereas covalent compounds are non-polar in nature. So they are insoluble in water.

d) Ionic compounds are usually hard crystals.
Due to strong attractive forces they are usually hard crystals.

Question 17.
If A, B, C and D are elements given that B is an inert gas (not Helium).

 Element Atomic Number A Z- 1 B Z C Z + 1 D Z + 4

What type of bonding would take place between (i) A and C and (ii) D and A and write their formulae?
1) Given that B is inert gas.
So the valency of A is 1 and it is a non-metal and valency of C is 1 and it is metal and valency of D is 4 and it is a non-metal.

2) The bond formed between C and A is ionic nature and its formula is CA.
The bond formed between D and A is covalent nature and its formula is DA4.

Question 18.
State whether the following statements are true or false. If statement is false, correct it.
i) In polar compounds the electropositive atom attracts the electron pair towards it.
This statement is wrong. In polar compounds the more electronegative atom attracts the electron pair towards it which is known as electronegativity.

ii) Hydrogen chloride gas is a di-polar molecule.
Chlorine is more electronegative than Hydrogen. So the electron pair shifts more towards chlorine atom. Thus partial positive charge is formed on Hydrogen and partial negative charge is formed on Chlorine. So it is dipolar molecule. So the statement is true.

iii) Covalent compounds are generally gases due to presence of weak electrostatic forces of attraction.
This statement is false. The forces are not electrostatic and also they are weak.

iv) Atoms achieve stable electronic configuration only by transfer of electrons from one atom to another.
This statement is false because they can acquire by not only electron transfer but also mutually sharing of electrons.

v) Ionic compounds are soft, solids or liquids at ordinary temperature.
This statement is false because ionic compounds are generally hard solids.

Question 19.
Five atoms are labelled from A to E.

 Atoms Atomic Number Mass Number A 20 40 B 9 19 C 3 7 D 8 16 E 7 14

a) Which one of these atoms (i) contain 7 protons, (ii) has an electronic configuration 2, 7?
i) The atom E contains 7 protons because its atomic number is 7.
ii) The atom with electronic configuration 2, 7 is B because its atomic number is 9.

b) Write down the formula of the compound formed between C and D.
The electronic configuration of C is 2, 1 since its atomic number is 3. The electronic configuration of D is 2, 6 since its atomic number is 8.
So the valency of C is 1. –
The valency of D is 2 (8 – 6 = 2).
∴ The compound formed between C and D is C2 D.

c) Predict which are (i) metals, (ii) non-metals?
Metals are A, C.
Non-metals are B, D, E.

Question 20.
Can you suggest an experiment to prove that ionic compounds possess strong bonds when compared to that of covalent bonds? Explain the procedure.

• Take a small amount of sodium chloride (NaCl) on a metal spatula (having an insulated handle).
• Heat it directly over the flame of a burner.
• We will see that sodium chloride (NaCl) does not melt easily.
• Sodium chloride melts (and becomes a liquid) only on strong heating.
• This shows that sodium chloride which is an ionic compound possesses strong bonds.
• So it has a high melting point.
• Whereas covalent compound like naphthalene and carbon tetra chloride has low boiling points such as 80° C and 77° C respectively.
• The force of attraction between the molecules of a covalent compound is very weak.
• Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compound has low melting points and low boiling points.

Question 21.
The electronic configurations of following elements are given below.

From these values complete the table.

 Compound Type of bonding Lithium chloride Lithium hydride Hydrogen chloride Carbon tetrachloride

a) Lithium is metal and chlorine is non-metal. So the bond is ionic in nature.
b) Lithium is metal and hydrogen is non-metal. So the bond is ionic in nature.
c) Hydrogen and chlorine are two dissimilar non-metals. So the bond formed is polar covalent bond.
d) Carbon and chlorine are dissimilar non-metals but electronegativity difference is less. So they form non-polar covalent bond.

 Compound Type of bonding Lithium chloride Ionic bond Lithium hydride Ionic bond Hydrogen chloride Polar covalent bond Carbon tetrachloride Non-polar covalent bond

Question 22.
Draw the structure of O2 by using valence bond theory.
O2 Formation :

Question 23.
Draw the structure of N2.

Question 24.
Draw the structure of molecules by Lewis method.
Formation of F2:

Formation of O2 :

Formation of N2 :

Formation of CH4 :

Formation of NH3 :

Formation of H2O :

Formation of BeCl2 :

Formation of BF3 :

Question 24.
Explain the formation of the following molecules using Lewis theory.
a) N2
b) O2
(OR)
Write the formation of double bond and triple bond according to Lewis theory.
a) Formation of N2 molecule by Lewis theory :

1. The electronic configuration of ’N’ atom is 2, 5 and to have octet in the valence shell it requires three more electrons.
2. When two nitrogen atoms approach each other, each atom contributes 3 electrons for bonding.
3. There are six electrons shared between two nitrogen atoms in the form of three pairs.
4. Therefore, there is a triple bond between two nitrogen atoms in N2 molecule.

b) Formation of O2 molecule by Lewis theory :

1. The electronic configuration of 8O is 2, 6.
2. Oxygen atom has six electrons in its valence shell.
3. It requires two more electrons to get octet in its valence shell.
4. Therefore oxygen atoms come close and each oxygen atom contributes two electrons for bonding.
5. Thus, there exist two covalent bonds between two oxygen atoms in O2 molecule as there are two pairs of electrons shared between them.
6. Two pairs of electrons are distributed between two oxygen atoms.
7. So, we can say that a double bond is formed between two oxygen atoms in O2 molecule.
8. By viewing the following diagram, both the oxygen atoms have octet in the valence shell.


text:
# All Questions

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text:
# Electrical conductor

In science and engineering, an electrical conductor is a material that contains movable electric charges. In metallic conductors, such as copper or aluminum, the movable charged particles are electrons. Positive charges may also be mobile in the form of atoms in a lattice missing electrons (called "holes") or ions, such as in the electrolyte of a battery.

## Details

The following applies to cases involving direct current only. When the direction of voltage/current alternates, other effects (inductance and capacitance) come into play also.

All conductors contain electric charges which will move when an electric potential difference (measured in volts) is applied across separate points on the material. This flow of charge (measured in amperes) is what is meant by electric current. In most materials, the rate of current is proportional to the voltage (Ohm's law,) provided the temperature remains constant and the material remains in the same shape and state. The ratio between the voltage and the current is called the resistance (measured in ohms) of the object between the points where the voltage was applied. The resistance across a standard mass (and shape) of a material at a given temperature is called the resistivity of the material. The inverse of resistance and resistivity is conductance and conductivity. Some good examples of conductors are metal.

Most familiar conductors are metallic. Copper is the most common material for electrical wiring (silver is the best but expensive), and gold for high-quality surface-to-surface contacts. However, there are also many non-metallic conductors, including graphite, solutions of salts, and all plasmas.

Non-conducting materials lack mobile charges, and so resist the flow of electric current, generating heat. In fact, all materials offer some resistance and warm up when a current flows. Thus, proper design of an electrical conductor takes into account the temperature that the conductor needs to be able to endure without damage, as well as the quantity of electrical current. The motion of charges also creates an electromagnetic field around the conductor that exerts a mechanical radial squeezing force on the conductor. A conductor of a given material and volume (length x cross-sectional area) has no real limit to the current it can carry without being destroyed as long as the heat generated by the resistive loss is removed and the conductor can withstand the radial forces. This effect is especially critical in printed circuits, where conductors are relatively small and close together, and inside an enclosure: the heat produced, if not properly removed, can cause fusing (melting) of the tracks.

Since all conductors have some resistance, and all insulators will carry some current, there is no theoretical dividing line between conductors and insulators. However, there is a large gap between the conductance of materials that will carry a useful current at working voltages and those that will carry a negligible current for the purpose in hand, so the categories of insulator and conductor do have practical utility.

Thermal and electrical conductivity often go together (for instance, most metals are both electrical and thermal conductors). However, some materials are practical electrical conductors but are not good thermal conductors.

## Electrical conductivity

Electrical conductivity or specific conductivity is a measure of a material's ability to conduct an electric current. When an electrical potential difference is placed across a conductor, its movable charges flow, giving rise to an electric current. The conductivity σ is defined as the ratio of the current density $\mathbf{J}$ to the electric field strength $\mathbf{E}$:

$\mathbf{J} = \sigma \mathbf{E}$

It is also possible to have materials in which the conductivity is anisotropic, in which case σ is a 3×3 matrix (or more technically a rank-2 tensor) which is generally symmetric.

Conductivity is the reciprocal (inverse) of electrical resistivity and has the SI units of siemens per meter (S·m-1), such as, if the electrical conductance between opposite faces of a 1-meter cube of material is 1 Siemens then the material's electrical conductivity is 1 Siemens per meter. Electrical conductivity is commonly represented by the Greek letter σ, but κ or γ are also occasionally used.

An EC meter is normally used to measure conductivity in a solution.

### Classification of materials by conductivity

Materials can be classified based on their electrical conductivity, as follows.

• A conductor such as a metal has high conductivity.
• An insulator like glass or a vacuum has low conductivity.
• The conductivity of a semiconductor is generally intermediate, but varies widely under different conditions, such as exposure of the material to electric fields or specific frequencies of light, and, most important, with temperature and composition of the semiconductor material.

The degree of doping in solid state semiconductors makes a large difference in conductivity. More doping leads to higher conductivity. The conductivity of a solution of water is highly dependent on its concentration of dissolved salts and sometimes other chemical species which tend to ionize in the solution. Electrical conductivity of water samples is used as an indicator of how salt-free or impurity-free the sample is; the purer the water, the lower the conductivity.

## Power engineering

In power engineering, a conductor is a piece of metal used to conduct electricity, known colloquially as an electrical wire.

### Conductor size

In many countries, conductors are measured by their cross section in square millimeters.

However, in the United States, conductors are measured by American wire gauge for smaller ones, and circular mils for larger ones. In some poor countries they have overloaded wires going into one circuit.

### Conductor materials

Of the metals commonly used for conductors, copper, has a high conductivity. Silver is more conductive, but due to cost it is not practical in most cases. However, it is used in specialized equipment, such as satellites, and as a thin plating to mitigate skin effect losses at high frequencies. Because of its ease of connection by soldering or clamping, copper is still the most common choice for most light-gauge wires. Aluminum has been used as a conductor in housing applications for cost reasons. It is actually more conductive than copper when compared by unit weight, but it has technical problems related to heat and compatibility of metals.

## Conductor voltage

The voltage on a conductor is determined by the connected circuitry and has nothing to do with the conductor itself. Conductors are usually surrounded by and/or supported by insulators and the insulation determines the maximum voltage that can be applied to any given conductor.

Voltage of a conductor "V" is given by

$V ={I}{R}$

where

I is the current, measured in amperes
V is the potential difference measured in volts
R is the resistance measured in ohms

## Conductor ampacity

The ampacity of a conductor, that is, the amount of current it can carry, is related to its electrical resistance: a lower-resistance conductor can carry more current. The resistance, in turn, is determined by the material the conductor is made from (as described above) and the conductor's size. For a given material, conductors with a larger cross-sectional area have less resistance than conductors with a smaller cross-sectional area.

For bare conductors, the ultimate limit is the point at which power lost to resistance causes the conductor to melt. Aside from fuses, most conductors in the real world are operated far below this limit, however. For example, household wiring is usually insulated with PVC insulation that is only rated to operate to about 60°C, therefore, the current flowing in such wires must be limited so that it never heats the copper conductor above 60°C, causing a risk of fire. Other, more expensive insulations such as Teflon or fiberglass may allow operation at much higher temperatures.

The American wire gauge article contains a table showing allowable ampacities for a variety of copper wire sizes.

## Isotropy

If an electric field is applied to a material, and the resulting induced electric current is in the same direction, the material is said to be an isotropic electrical conductor. If the resulting electric current is in a different direction from the applied electric field, the material is said to be an anisotropic electrical conductor!.