GUJCET 2007 Chemistry Question Paper with Answer and Solution

(D) Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$.
Since the density $\rho$ is the same,the mass $m$ is proportional to the volume $V = \frac{4}{3} \pi r^3$.
Therefore,the ratio of masses is $\frac{m_1}{m_2} = \frac{r_1^3}{r_2^3} = (\frac{1}{2})^3 = \frac{1}{8}$.
Since the nucleus is initially at rest,the total momentum is zero. According to the law of conservation of linear momentum:
$m_1 v_1 + m_2 v_2 = 0$
$m_1 v_1 = -m_2 v_2$
Taking the magnitude,$|\frac{v_1}{v_2}| = \frac{m_2}{m_1} = \frac{8}{1}$.
Thus,the ratio of their velocities is $8 : 1$.

(C) The given reaction is the free radical chlorination of $n$-butane.
This reaction proceeds via a free radical mechanism initiated by ultraviolet $(UV)$ light.
$CH_3-CH_2-CH_2-CH_3 + Cl_2 \xrightarrow{UV \ light} CH_3-CH_2-CH_2-CH_2-Cl + CH_3-CH_2-CHCl-CH_3 + HCl$.
Therefore,the correct reagent is $Cl_2 / UV$ light.

(B) According to the principles of electrostatics,a charge at rest produces an electric field in the space surrounding it.
It does not produce a magnetic field because a magnetic field is generated only by moving charges (currents).
Therefore,a stationary charge produces an electric field only.
Thus,the correct option is $B$.

(A) The general formula for the unit of the rate constant $K$ for a reaction of order $n$ is given by:
$Unit = \left(\frac{mol}{L}\right)^{1-n} \cdot s^{-1}$
For a third-order reaction,$n = 3$.
Substituting $n = 3$ into the formula:
$Unit = \left(\frac{mol}{L}\right)^{1-3} \cdot s^{-1} = \left(\frac{mol}{L}\right)^{-2} \cdot s^{-1}$
This can also be written as:
$\left(\frac{L}{mol}\right)^2 \cdot s^{-1}$
Comparing this with the given options,option $A$ is $\left(\frac{L}{mol}\right)^2 \cdot s^{-1}$,which matches our derived result.
Therefore,the correct option is $A$.

(D) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
For $Fe^{3+}$,the configuration is $[Ar] 3d^5$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
In $Fe^{3+}$ $(3d^5)$,the five electrons are paired such that four electrons form two pairs and one electron remains unpaired $(n=1)$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(C) Geometrical isomerism is shown by complexes that have different spatial arrangements of ligands around the central metal ion.
For octahedral complexes of the type $[MA_6]$,$[MA_5B]$,$[MA_4B_2]$,$[MA_3B_3]$,and $[MA_2B_4]$,geometrical isomerism is possible.
Option $A$: $[Co(NH_3)_4Cl_2]^+$ is of the type $[MA_4B_2]$,which shows $cis$ and $trans$ isomers.
Option $B$: $[Fe(NH_3)_2(CN)_4]^-$ is of the type $[MA_2B_4]$,which shows $cis$ and $trans$ isomers.
Option $C$: $[Cr(OX)_3]^{3-}$ contains a bidentate ligand $(OX^{2-})$. Complexes of the type $[M(AA)_3]$ do not show geometrical isomerism because all positions are equivalent.
Option $D$: $[Co(NH_3)_3(NO_2)_3]$ is of the type $[MA_3B_3]$,which shows facial $(fac)$ and meridional $(mer)$ isomers.
Therefore,the correct answer is $C$.

(A) The maximum oxidation state of a transition element is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.
For option $A$: $(n-1)d^5 ns^2$,total electrons = $5 + 2 = 7$.
For option $B$: $(n-1)d^8 ns^2$,total electrons = $8 + 2 = 10$ (but the maximum stable oxidation state is usually limited by the number of unpaired electrons and stability of the configuration).
For option $C$: $(n-1)d^5 ns^1$,total electrons = $5 + 1 = 6$.
For option $D$: $(n-1)d^3 ns^2$,total electrons = $3 + 2 = 5$.
Among the given configurations,the configuration $(n-1)d^5 ns^2$ (e.g.,$Mn$ in $3d^5 4s^2$) allows for a maximum oxidation state of $+7$. While $d^8 s^2$ has more electrons,it does not exhibit an oxidation state of $+10$; transition metals typically show a maximum oxidation state corresponding to the sum of $ns$ and $(n-1)d$ electrons up to $Mn$ $(+7)$. Thus,$(n-1)d^5 ns^2$ is the correct choice for the highest common oxidation state.

(A) The $d-d$ transition is possible only in ions that have partially filled $d$-orbitals ($d^1$ to $d^9$ configuration).
$1$. $Ti^{4+}$: The electronic configuration is $[Ar] 3d^0$. Since there are no electrons in the $d$-orbital,$d-d$ transition is not possible.
$2$. $Cr^{3+}$: The electronic configuration is $[Ar] 3d^3$. It has partially filled $d$-orbitals,so $d-d$ transition is possible.
$3$. $Mn^{2+}$: The electronic configuration is $[Ar] 3d^5$. It has partially filled $d$-orbitals,so $d-d$ transition is possible.
$4$. $Cu^{2+}$: The electronic configuration is $[Ar] 3d^9$. It has partially filled $d$-orbitals,so $d-d$ transition is possible.
Therefore,the correct option is $A$.