(A) : As we know that the crystal field stabilisation energy is inversely proportional to the wavelength of absorption of light. The compounds are - \(\begin{array}{ccc}{\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-}} & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\ +2 & +2 & +2\end{array}\) All the compounds have same metal with same oxidation state then we move to the nature of ligand, according to the spectrochemical series, the placing order of ligands are $\mathrm{H}_{2} \mathrm{O}<\mathrm{NH}_{3}<\mathrm{NO}_{2}$. Thus, the CFSE order of the compounds are $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)\right]^{4-}$ Hence, the order of wavelength of absorption will be $\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$.
AIIMS-2005
COORDINATION COMPOUNDS
274320
The number of unpaired electrons in $\mathrm{Ni}(\mathrm{CO})_{4}$ is-
1 0
2 1
3 3
4 4
Explanation:
(A) : $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $x+4(0)=0$ $x=0$ $\mathrm{CO}$ is a strong field ligand thus pairing occur in inner orbitals. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ four sp hybrid orbital The number of unpaired electron is 0 .
BCECE-2007
COORDINATION COMPOUNDS
274326
Crystal field stabilization energy for high spin $\mathrm{d}^{4}$ octahedral complex is
274327
Low spin complex of $d^{6}$-cation in an octahedral field will have the following energy
1 $\frac{-12}{5} \Delta_{0}+P$
2 $\frac{-12}{5} \Delta_{0}+3 \mathrm{P}$
3 $\frac{-2}{5} \Delta_{0}+2 \mathrm{P}$
4 $\frac{-2}{5} \Delta_{0}+P$ $\left(\Delta_{0}=\right.$ crystal field splitting energy in an octahedral field, $\mathrm{P}=$ Electron pairing energy)
Explanation:
(B) : For octahedral d ${ }^{6}$ (low spin)- CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}+\mathrm{mp}$ Where $-\mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}=$ no. of electron in $\mathrm{t}_{2 \mathrm{~g}}$ orbital $\mathrm{N}_{\mathrm{e}_{\mathrm{g}}}=$ no. of electron in $\mathrm{t}_{\mathrm{e}_{\mathrm{g}}}$ orbital $\mathrm{mp}=$ no. of forcefully paired electron. CFSE $=[-0.4 \times 6+0.6 \times 0] \Delta_{\mathrm{o}}+3 \mathrm{P}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}+3 \mathrm{P}$
(A) : As we know that the crystal field stabilisation energy is inversely proportional to the wavelength of absorption of light. The compounds are - \(\begin{array}{ccc}{\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-}} & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\ +2 & +2 & +2\end{array}\) All the compounds have same metal with same oxidation state then we move to the nature of ligand, according to the spectrochemical series, the placing order of ligands are $\mathrm{H}_{2} \mathrm{O}<\mathrm{NH}_{3}<\mathrm{NO}_{2}$. Thus, the CFSE order of the compounds are $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)\right]^{4-}$ Hence, the order of wavelength of absorption will be $\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$.
AIIMS-2005
COORDINATION COMPOUNDS
274320
The number of unpaired electrons in $\mathrm{Ni}(\mathrm{CO})_{4}$ is-
1 0
2 1
3 3
4 4
Explanation:
(A) : $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $x+4(0)=0$ $x=0$ $\mathrm{CO}$ is a strong field ligand thus pairing occur in inner orbitals. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ four sp hybrid orbital The number of unpaired electron is 0 .
BCECE-2007
COORDINATION COMPOUNDS
274326
Crystal field stabilization energy for high spin $\mathrm{d}^{4}$ octahedral complex is
274327
Low spin complex of $d^{6}$-cation in an octahedral field will have the following energy
1 $\frac{-12}{5} \Delta_{0}+P$
2 $\frac{-12}{5} \Delta_{0}+3 \mathrm{P}$
3 $\frac{-2}{5} \Delta_{0}+2 \mathrm{P}$
4 $\frac{-2}{5} \Delta_{0}+P$ $\left(\Delta_{0}=\right.$ crystal field splitting energy in an octahedral field, $\mathrm{P}=$ Electron pairing energy)
Explanation:
(B) : For octahedral d ${ }^{6}$ (low spin)- CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}+\mathrm{mp}$ Where $-\mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}=$ no. of electron in $\mathrm{t}_{2 \mathrm{~g}}$ orbital $\mathrm{N}_{\mathrm{e}_{\mathrm{g}}}=$ no. of electron in $\mathrm{t}_{\mathrm{e}_{\mathrm{g}}}$ orbital $\mathrm{mp}=$ no. of forcefully paired electron. CFSE $=[-0.4 \times 6+0.6 \times 0] \Delta_{\mathrm{o}}+3 \mathrm{P}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}+3 \mathrm{P}$
(A) : As we know that the crystal field stabilisation energy is inversely proportional to the wavelength of absorption of light. The compounds are - \(\begin{array}{ccc}{\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-}} & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\ +2 & +2 & +2\end{array}\) All the compounds have same metal with same oxidation state then we move to the nature of ligand, according to the spectrochemical series, the placing order of ligands are $\mathrm{H}_{2} \mathrm{O}<\mathrm{NH}_{3}<\mathrm{NO}_{2}$. Thus, the CFSE order of the compounds are $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)\right]^{4-}$ Hence, the order of wavelength of absorption will be $\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$.
AIIMS-2005
COORDINATION COMPOUNDS
274320
The number of unpaired electrons in $\mathrm{Ni}(\mathrm{CO})_{4}$ is-
1 0
2 1
3 3
4 4
Explanation:
(A) : $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $x+4(0)=0$ $x=0$ $\mathrm{CO}$ is a strong field ligand thus pairing occur in inner orbitals. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ four sp hybrid orbital The number of unpaired electron is 0 .
BCECE-2007
COORDINATION COMPOUNDS
274326
Crystal field stabilization energy for high spin $\mathrm{d}^{4}$ octahedral complex is
274327
Low spin complex of $d^{6}$-cation in an octahedral field will have the following energy
1 $\frac{-12}{5} \Delta_{0}+P$
2 $\frac{-12}{5} \Delta_{0}+3 \mathrm{P}$
3 $\frac{-2}{5} \Delta_{0}+2 \mathrm{P}$
4 $\frac{-2}{5} \Delta_{0}+P$ $\left(\Delta_{0}=\right.$ crystal field splitting energy in an octahedral field, $\mathrm{P}=$ Electron pairing energy)
Explanation:
(B) : For octahedral d ${ }^{6}$ (low spin)- CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}+\mathrm{mp}$ Where $-\mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}=$ no. of electron in $\mathrm{t}_{2 \mathrm{~g}}$ orbital $\mathrm{N}_{\mathrm{e}_{\mathrm{g}}}=$ no. of electron in $\mathrm{t}_{\mathrm{e}_{\mathrm{g}}}$ orbital $\mathrm{mp}=$ no. of forcefully paired electron. CFSE $=[-0.4 \times 6+0.6 \times 0] \Delta_{\mathrm{o}}+3 \mathrm{P}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}+3 \mathrm{P}$
(A) : As we know that the crystal field stabilisation energy is inversely proportional to the wavelength of absorption of light. The compounds are - \(\begin{array}{ccc}{\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-}} & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\ +2 & +2 & +2\end{array}\) All the compounds have same metal with same oxidation state then we move to the nature of ligand, according to the spectrochemical series, the placing order of ligands are $\mathrm{H}_{2} \mathrm{O}<\mathrm{NH}_{3}<\mathrm{NO}_{2}$. Thus, the CFSE order of the compounds are $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)\right]^{4-}$ Hence, the order of wavelength of absorption will be $\left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+2}<\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$.
AIIMS-2005
COORDINATION COMPOUNDS
274320
The number of unpaired electrons in $\mathrm{Ni}(\mathrm{CO})_{4}$ is-
1 0
2 1
3 3
4 4
Explanation:
(A) : $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $x+4(0)=0$ $x=0$ $\mathrm{CO}$ is a strong field ligand thus pairing occur in inner orbitals. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ four sp hybrid orbital The number of unpaired electron is 0 .
BCECE-2007
COORDINATION COMPOUNDS
274326
Crystal field stabilization energy for high spin $\mathrm{d}^{4}$ octahedral complex is
274327
Low spin complex of $d^{6}$-cation in an octahedral field will have the following energy
1 $\frac{-12}{5} \Delta_{0}+P$
2 $\frac{-12}{5} \Delta_{0}+3 \mathrm{P}$
3 $\frac{-2}{5} \Delta_{0}+2 \mathrm{P}$
4 $\frac{-2}{5} \Delta_{0}+P$ $\left(\Delta_{0}=\right.$ crystal field splitting energy in an octahedral field, $\mathrm{P}=$ Electron pairing energy)
Explanation:
(B) : For octahedral d ${ }^{6}$ (low spin)- CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}+\mathrm{mp}$ Where $-\mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}=$ no. of electron in $\mathrm{t}_{2 \mathrm{~g}}$ orbital $\mathrm{N}_{\mathrm{e}_{\mathrm{g}}}=$ no. of electron in $\mathrm{t}_{\mathrm{e}_{\mathrm{g}}}$ orbital $\mathrm{mp}=$ no. of forcefully paired electron. CFSE $=[-0.4 \times 6+0.6 \times 0] \Delta_{\mathrm{o}}+3 \mathrm{P}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}+3 \mathrm{P}$
