IIT JEE 1989 Chemistry Question Paper with Answer and Solution

(A) The atomic number of chromium $(Cr)$ is $24$.
According to the Aufbau principle,the expected configuration is $[Ar] \, 3d^4 \, 4s^2$.
However,a $3d$ subshell that is half-filled $(3d^5)$ is more stable due to exchange energy and symmetry.
Therefore,one electron from the $4s$ orbital jumps into the $3d$ orbital to achieve the stable configuration $[Ar] \, 3d^5 \, 4s^1$.

(B) $BF_3$ (Boron trifluoride) has a trigonal planar geometry where the three $B-F$ bond dipoles cancel each other out,resulting in a net dipole moment of $0 \ D$.

(B) The structure of $but-1-ene-3-yne$ is $CH_2=CH-C \equiv CH$.
Counting the bonds:
Sigma $( \sigma )$ bonds: There are $3$ $C-H$ bonds,$1$ $C-C$ single bond,$1$ $C=C$ double bond (contains $1$ $ \sigma $),and $1$ $C \equiv C$ triple bond (contains $1$ $ \sigma $). Total $ \sigma $ bonds = $3 + 1 + 1 + 1 + 1 = 7$.
Pi $( \pi )$ bonds: The $C=C$ bond contains $1$ $ \pi $ bond,and the $C \equiv C$ bond contains $2$ $ \pi $ bonds. Total $ \pi $ bonds = $1 + 2 = 3$.
Thus,the molecule has $7$ $ \sigma $ bonds and $3$ $ \pi $ bonds.

(C) In the compound $CH_3-CH_2-OH$,the carbon atom marked $C^*$ (the one attached to the hydroxyl group) is bonded to four other atoms (three hydrogens and one oxygen) via single bonds.
Since it forms four sigma bonds and has no lone pairs,its steric number is $4$,which corresponds to $sp^3$ hybridization.
In $H-COOH$,$(NH_2)_2-C=O$,and $CH_3-CHO$,the carbon atom $C^*$ is involved in a double bond with oxygen,making it $sp^2$ hybridized.

(A) In $PCl_3$,the central $P$ atom undergoes $sp^3$ hybridization.
It has $3$ bond pairs and $1$ lone pair of electrons.
According to $VSEPR$ theory,the presence of one lone pair causes the geometry to be pyramidal,similar to $NH_3$.

(C) According to Molecular Orbital Theory $(MOT)$,a species is paramagnetic if it contains one or more unpaired electrons.
$1$. $O_2^-$ has $17$ electrons. Its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Since it has an unpaired electron,it is paramagnetic.
$2$. $NO$ has $15$ electrons. Its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Since it has an unpaired electron,it is paramagnetic.
$3$. $CN^-$ has $14$ electrons. All electrons are paired,so it is diamagnetic.
Therefore,both $(a)$ and $(b)$ are paramagnetic.

(B) In an ideal gas,there are no intermolecular forces of attraction between the particles.
During unrestrained expansion (expansion against vacuum),no work is done by the gas $(w = 0)$.
Since there are no attractive forces to overcome,no internal energy is consumed to separate the molecules,and therefore,the temperature of the ideal gas remains constant.

(C) The Van der Waals constant '$a$' is a measure of the magnitude of the attractive forces between the gas molecules.
Greater the value of '$a$',stronger are the intermolecular forces of attraction.
Stronger intermolecular forces make it easier for a gas to be liquefied.
Comparing the given values: $a(NH_3) = 4.170 > a(CH_4) = 2.253 > a(N_2) = 1.390 > a(O_2) = 1.3$.
Since $NH_3$ has the highest value of '$a$',it can be most easily liquefied.

(B) The reaction $N_{2(g)} + O_{2(g)} \to 2NO_{(g)}$ is an endothermic reaction because heat is absorbed during the process.
All other reactions listed,such as combustion and neutralization,are exothermic reactions where heat is released.

(D) . $Na^{+} < F^{-} < O^{2-} < N^{3-}$
All these ions are isoelectronic,meaning they contain the same number of electrons ($10$ electrons).
For isoelectronic species,the ionic radius decreases as the atomic number $(Z)$ increases because the effective nuclear charge increases.
Since the atomic numbers are $N(7)$,$O(8)$,$F(9)$,and $Na(11)$,the effective nuclear charge is highest for $Na^{+}$,resulting in the smallest ionic size.

(C) The solubility of an ionic compound in water depends on the balance between its lattice energy and hydration energy.
For a salt to be soluble,the hydration energy must be greater than the lattice energy.
In the case of $Na_2SO_4$,the hydration energy is greater than the lattice energy,making it soluble.
In the case of $BaSO_4$,the lattice energy is significantly higher than the hydration energy due to the high charge density and strong ionic interaction,which makes it sparingly soluble.

(D) An isopropyl group is represented by the structure $(CH_3)_2CH-$.
$1.$ $2, 2, 3, 3-$tetramethylpentane: All methyl groups are attached to quaternary carbons,so no isopropyl group is present.
$2.$ $2, 2-$dimethylpentane: The structure is $CH_3-CH_2-CH_2-C(CH_3)_2-CH_3$. It contains a tert-butyl group,not an isopropyl group.
$3.$ $2, 2, 3-$trimethylpentane: The structure is $CH_3-C(CH_3)_2-CH(CH_3)-CH_2-CH_3$. It contains a tert-butyl group and a sec-butyl group,but no isopropyl group.
$4.$ $2-$methylpentane: The structure is $(CH_3)_2CH-CH_2-CH_2-CH_3$. This compound contains one isopropyl group at the $2-$position.

(C) In $(CH_3)_3COH$,the structure is $(CH_3)_3C-OH$.
All the carbon atoms are bonded to four other atoms via single bonds,meaning they are all $sp^3$ hybridized.
In $HCOOH$,the carbonyl carbon is $sp^2$ hybridized.
In $(NH_2)_2CO$,the carbonyl carbon is $sp^2$ hybridized.
In $(CH_3)_3CHO$,the carbonyl carbon is $sp^2$ hybridized.

(D) An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
In $CH_3-CH(OH)-CH_2Br$,the second carbon atom is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2Br$.
Therefore,it is an asymmetric carbon atom.

(C) Bond length is inversely proportional to the $s$-character of the hybrid orbital of the carbon atom.
$1.$ In $C_2H_2$ (Ethyne),the hybridization of carbon is $sp$,which has $50 \% \ s$-character.
$2.$ In $C_2H_4$ (Ethene),the hybridization of carbon is $sp^2$,which has $33.33 \% \ s$-character.
$3.$ In $C_2H_6$ (Ethane),the hybridization of carbon is $sp^3$,which has $25 \% \ s$-character.
Since $C_2H_6$ has the lowest $s$-character $(25 \%)$,the $C-H$ bond length is the greatest in $C_2H_6$.

(A) The acidic nature of hydrogen in hydrocarbons depends on the hybridization of the carbon atom to which it is attached.
In $CH \equiv CH$ (Ethyne),the carbon is $sp$ hybridized,which has $50\%$ $s$-character,making it more electronegative and thus the $C-H$ bond is polar,allowing the hydrogen to be acidic.
Reaction with sodium metal confirms this: $2CH \equiv CH + 2Na \to 2CH \equiv C^{-}Na^{+} + H_2 \uparrow$.
Therefore,ethyne has acidic hydrogen.

(C) The atomic number of chlorine $(Cl)$ is $17$.
The electronic configuration of $Cl$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.
The valence shell configuration is $3s^2 3p^5$.
In the $3p$ subshell,there are $3$ orbitals $(3p_x, 3p_y, 3p_z)$. According to Hund's rule,the $5$ electrons are filled as follows: $3p_x^2, 3p_y^2, 3p_z^1$.
The unpaired electron is in the $3p_z$ orbital.
For this electron:
Principal quantum number $(n)$ = $3$
Azimuthal quantum number $(l)$ for $p$-orbital = $1$
Magnetic quantum number $(m)$ can be $-1, 0, +1$. By convention,the unpaired electron is often assigned $m = +1$.
Thus,the set of quantum numbers is $n=3, l=1, m=1$.

(A) Let the mass of both gases be $m \ g$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{m}{16}$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{m}{2}$.
The partial pressure of a gas is proportional to its mole fraction $(x_i = \frac{n_i}{n_{total}})$.
Mole fraction of $H_2$ $(x_{H_2})$ = $\frac{n_{H_2}}{n_{H_2} + n_{CH_4}} = \frac{\frac{m}{2}}{\frac{m}{2} + \frac{m}{16}}$.
$x_{H_2} = \frac{\frac{m}{2}}{\frac{8m + m}{16}} = \frac{\frac{m}{2}}{\frac{9m}{16}} = \frac{m}{2} \times \frac{16}{9m} = \frac{8}{9}$.
Therefore,the partial pressure caused by $H_2$ is $\frac{8}{9}$ of the total pressure.

(D) An isopropyl group is represented by the structure $(CH_3)_2CH-$.
$1.$ $2,2,3,3-$tetramethylpentane: The structure is $(CH_3)_3C-C(CH_3)_2-CH_2-CH_3$. It does not contain an isopropyl group.
$2.$ $2,2-$dimethylpentane: The structure is $(CH_3)_3C-CH_2-CH_2-CH_3$. It does not contain an isopropyl group.
$3.$ $2,2,3-$trimethylpentane: The structure is $(CH_3)_3C-CH(CH_3)-CH_2-CH_3$. It contains a tert-butyl group,but no isopropyl group.
$4.$ $2-$methylpentane: The structure is $(CH_3)_2CH-CH_2-CH_2-CH_3$. This molecule contains one isopropyl group at the end of the chain.

(D) The thermal stress $\sigma$ developed in a rod fixed between two rigid walls when heated by a temperature change $\Delta T$ is given by the formula: $\sigma = Y \alpha \Delta T$.
Given that the thermal stress in both rods is equal,we have: $\sigma_1 = \sigma_2$.
Substituting the formula for thermal stress: $Y_1 \alpha_1 \Delta T = Y_2 \alpha_2 \Delta T$.
Since the temperature change $\Delta T$ is the same for both rods,we can cancel it from both sides: $Y_1 \alpha_1 = Y_2 \alpha_2$.
Rearranging the terms to find the ratio $Y_1 : Y_2$: $\frac{Y_1}{Y_2} = \frac{\alpha_2}{\alpha_1}$.
Given the ratio $\alpha_1 : \alpha_2 = 2 : 3$,we have $\frac{\alpha_1}{\alpha_2} = \frac{2}{3}$,which implies $\frac{\alpha_2}{\alpha_1} = \frac{3}{2}$.
Therefore,$\frac{Y_1}{Y_2} = \frac{3}{2}$.

(C) The work function $\Phi = 4.2 \, eV = 4.2 \times 1.602 \times 10^{-19} \, J \approx 6.73 \times 10^{-19} \, J$.
The energy of the incident radiation $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2000 \times 10^{-10}} \, J = 9.939 \times 10^{-19} \, J$.
According to Einstein's photoelectric equation,$K_{max} = E - \Phi$.
$K_{max} = (9.939 - 6.73) \times 10^{-19} \, J = 3.209 \times 10^{-19} \, J$.
Thus,the kinetic energy is approximately $3.2 \times 10^{-19} \, J$.

(A) The correct answer is $(A)$.
In the group $15$ hydrides ($NH_3$ to $SbH_3$),the basic strength decreases down the group.
This is because the lone pair on the central atom is present in a larger orbital as the size of the central atom increases.
In $NH_3$,the lone pair is in a smaller $sp^3$ hybridized orbital,making it more concentrated and easily available for donation compared to the larger atoms in $PH_3$,$AsH_3$,and $SbH_3$.

(A) Nitrogen $(I)$ oxide,also known as nitrous oxide $(N_2O)$,is prepared by the thermal decomposition of ammonium nitrate $(NH_4NO_3)$.
The chemical equation is:
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
Therefore,the correct option is $A$.

(A) The oxidation states of chlorine in the given oxoacids are as follows:
$HClO_4$: $+7$
$HClO_3$: $+5$
$HClO_2$: $+3$
$HClO$: $+1$
As the oxidation number of the central halogen atom increases,the acidic strength of the oxoacid increases.
Therefore,the order of acidic strength is $HClO_4 > HClO_3 > HClO_2 > HClO$.
Thus,$HClO_4$ is the strongest acid.

(C) Concentrated $HNO_3$ acts as a strong oxidizing agent and oxidizes iodine $(I_2)$ to iodic acid $(HIO_3)$.
The balanced chemical equation for the reaction is:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$
Therefore,the correct option is $C$.

(D) The depression in freezing point is a colligative property,which depends on the number of particles (ions) produced in the solution.
$1$. $NaCl \to Na^{+} + Cl^{-}$ ($2$ ions)
$2$. Urea is a non-electrolyte ($1$ particle)
$3$. Glucose is a non-electrolyte ($1$ particle)
$4$. $K_2SO_4 \to 2K^{+} + SO_4^{2-}$ ($3$ ions)
Since $K_2SO_4$ produces the maximum number of ions ($3$ ions) per formula unit,it will exhibit the maximum depression in freezing point.

(B) The half-life $(t_{1/2})$ of a radioactive sample is given by the formula $t_{1/2} = \frac{\ln 2}{\lambda}$.
The mean life $(\tau)$ of a radioactive sample is the reciprocal of the decay constant,given by $\tau = \frac{1}{\lambda}$.
Therefore,the half-life and mean life are $\frac{\ln 2}{\lambda}$ and $\frac{1}{\lambda}$ respectively.

(A) The froth flotation process is primarily used for the concentration of sulphide ores.
$A$ Chalcopyrite $(CuFeS_2)$ is a sulphide ore,which is why it is concentrated by the froth flotation process.