IIT JEE 1979 Chemistry Question Paper with Answer and Solution

(B) The atomic number of zinc $(Zn)$ is $30$.
For the isotope with mass number $A = 70$,the number of neutrons is calculated as $N = A - Z$.
$N = 70 - 30 = 40$.
The formation of a dipositive ion $(Zn^{2+})$ involves the loss of electrons from the valence shell,which does not affect the number of protons or neutrons in the nucleus.
Therefore,the number of neutrons remains $40$.

(A) The atomic number of $C$ is $6$ and the atomic number of $O$ is $8$.
In one molecule of $CO_2$,there is $1$ atom of $C$ and $2$ atoms of $O$.
Total number of electrons = $(1 \times 6) + (2 \times 8) = 6 + 16 = 22$ electrons.
Therefore,the correct option is $A$.

(B) The octet rule states that atoms tend to gain,lose,or share electrons to achieve a stable configuration of $8$ valence electrons.
In $CO_2$,$H_2O$,and $O_2$,all atoms (except $H$ in $H_2O$) achieve an octet.
In $NO$ (Nitric oxide),the nitrogen atom has $5$ valence electrons and the oxygen atom has $6$ valence electrons,totaling $11$ valence electrons.
Since the total number of valence electrons is odd,it is impossible for all atoms to satisfy the octet rule,making $NO$ an odd-electron molecule.

(A) The $Law \ of \ mass \ action$ states that the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reacting substances.
Active mass is defined as the molar concentration of the substance,i.e.,$Active \ mass = [Concentration]$.

(C) In $KCN$,the bond between $K^+$ and $CN^-$ is ionic.
Within the cyanide ion $(CN^-)$,there is a triple covalent bond between the carbon and nitrogen atoms $(C \equiv N)$.
Therefore,$KCN$ contains both ionic and covalent bonds.

(C) In steady state, capacitors act as open circuits. The current flows only through the resistors. The total resistance in the circuit is $R_{eq} = 20\, \Omega + 10\, \Omega = 30\, \Omega$. The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{100\, V}{30\, \Omega} = \frac{10}{3}\, A$.
The potential at point $A$ is $V_A$ and at point $C$ is $V_C$. The potential difference $V_{AC} = V_A - V_C = 100\, V$.
Let the potential at point $B$ be $V_B$. The capacitors form a bridge-like structure. The equivalent capacitance between $A$ and $B$ is $C_{AB} = 3\, \mu F + 3\, \mu F = 6\, \mu F$ (parallel combination). The equivalent capacitance between $B$ and $C$ is $C_{BC} = 1\, \mu F + 1\, \mu F = 2\, \mu F$ (parallel combination).
Using the principle of voltage division for capacitors in series (since the charge $Q$ on the branches $AB$ and $BC$ must be equal in steady state), we have $V_{AB} = V_{total} \cdot \frac{C_{BC}}{C_{AB} + C_{BC}} = 100 \cdot \frac{2}{6+2} = 100 \cdot \frac{2}{8} = 25\, V$.
Similarly, $V_{BC} = V_{total} \cdot \frac{C_{AB}}{C_{AB} + C_{BC}} = 100 \cdot \frac{6}{6+2} = 100 \cdot \frac{6}{8} = 75\, V$.
Thus, $V_{AB} = 25\, V$ and $V_{BC} = 75\, V$.

(C) Nitric oxide $(NO)$ reacts with atmospheric oxygen to form nitrogen dioxide $(NO_2)$.
The chemical equation is:
$2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)$
Nitrogen dioxide $(NO_2)$ is a reddish-brown coloured gas.