Q9 / 160 · TS EAMCET 2014 · Chemistry
Match the following.
$\begin{array}{llll}A & B & C & D\end{array}$
$\begin{array}{llll}A & B & C & D\end{array}$
- A(v)$\quad$(ii)$\quad$(iv)$\quad$(iii)
- Correct answerB(ii)$\quad$(iii)$\quad$(iv)$\quad$(v)
- C(ii)$\quad$ (iii)$\quad$ (i) $\quad$ (v)
- D(iii)$\quad$ (ii)$\quad$ (iv)$\quad$ (i)
Solution
I. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
Here Co is present as $\mathrm{Co}^{3+}$ ion. $\mathrm{NH}_3$ being strong field ligand pair up the unpaired electrons of $\mathrm{Co}$.
II.$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$
Here $\mathrm{Ni}$ is present in its ground state. $\mathrm{CO}$ being a strong field ligand, pair up the unpaired electrons of $\mathrm{Ni}$
$$
{ }_{28} \mathrm{Ni}=[\mathrm{Ar}] 3 d^8 4 s^2
$$
III. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
Here $\mathrm{Pt}$ is present as $\mathrm{Pt}^{2+}$ ion.
$$
\begin{aligned}
{ }_{78} \mathrm{Pt} & =[\mathrm{Xe}] 5 d^9 6 s^1 \\
\mathrm{Pt}^{2+} & =[\mathrm{Xe}] 5 d^8 6 \mathrm{~s}^0
\end{aligned}
$$
IV.$\left[\mathrm{CoF}_6{ }^{\beta-}\right.]$ Here $\mathrm{Co}$ is present as $\mathrm{Co}^{3+}$ ion, $\mathrm{F}$ being weak field ligand is unable to pair up its unpaired electrons. $$ \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6 4 s^0 $$ V.$\left[\mathrm{Fe}(\mathrm{CO})_5\right]$ Here $\mathrm{Fe}$ is present in its ground state. $$ { }_{26} \mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2 $$
IV.$\left[\mathrm{CoF}_6{ }^{\beta-}\right.]$ Here $\mathrm{Co}$ is present as $\mathrm{Co}^{3+}$ ion, $\mathrm{F}$ being weak field ligand is unable to pair up its unpaired electrons. $$ \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6 4 s^0 $$ V.$\left[\mathrm{Fe}(\mathrm{CO})_5\right]$ Here $\mathrm{Fe}$ is present in its ground state. $$ { }_{26} \mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2 $$
Hence, the correct answer is option: 'B'.