274141
Which of the following outer octahedral complexes have same number of unpaired electrons? 1.$\left(\mathrm{MnCl}_{6}\right)^{3-}$ 3.$\left(\mathrm{CoF}_{6}\right)^{3-}$ 2.$\left(\mathrm{FeF}_{6}\right)^{3-}$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$
1 1 and 3
2 1 and 2
3 3 and 4
4 2 and 3
Explanation:
(A) : All four complexes given are outer orbital or high spin octahedral complexes. So, d-orbital electrons of central metal ions of the complexes remain unaffected by weak field ligands. 1.$\left[\mathrm{MnCl}_{6}\right]^{3-} \rightarrow \mathrm{Mn}^{3+} \Rightarrow \mathrm{d}^{4}, \mathrm{n}=4$ $[\because \mathrm{n}=$ Number of unpaired electrons $]$ 2.$\left[\mathrm{FeF}_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+} \Rightarrow \mathrm{d}^{5}, \mathrm{n}=5$ 3.$\left[\mathrm{CoF}_{6}\right]^{3-} \rightarrow \mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}, \mathrm{n}=4$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} \rightarrow \mathrm{Ni}^{2+} \Rightarrow \mathrm{d}^{8}, \mathrm{n}=2$ So, complexes 1 and 3 have same number of unpaired electrons $(\mathrm{n}=4)$.
AP EAMCET (Engg.) 17.09.2020
COORDINATION COMPOUNDS
274142
The ion that has $\mathbf{s p}^{3} \mathrm{~d}^{2}$ hybridization for the central atom,
274144
The complexes $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are, respectively
1 tetrahedral and square planar
2 tetrahedral and tetrahedral
3 square planar and square planar
4 square planar and tetrahedral.
Explanation:
(A) $\text { : } {\left[\mathrm{Ni}(\mathrm{CO})_4\right] }$ $x+4(0)=0$ $x=0$ Pairing of electron takes place because CO is SFL. Pairing of electron takes place because $\mathrm{CO}$ is $\mathrm{SFL}$. $\mathrm{Ni}=[\operatorname{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2 4 \mathrm{p}^0$ Hence, the complexes $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are tetrahedral and square planar respectively And, $\quad\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ $x+4(-1)=-2$ $x=+2$ Pairing of electron takes place because CN is SFL. $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
AMU-2019
COORDINATION COMPOUNDS
274147
Regarding $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{--}$complex ion which of the following statements is true?
1 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is paramagnetic.
2 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is diamagnetic
3 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is diamagnetic
4 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is paramagnetic.
Explanation:
(B) : $\left(\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $\quad \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 \text { (Pairing of electron occur because } \mathrm{CN}$ $\text { is SFL). }$
Tripura JEE-2019
COORDINATION COMPOUNDS
274148
Which of the given options are correct for $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ complex?
1 $\mathrm{d}^2 \mathrm{sp}^2$ Hybridization
2 $\mathrm{sp}^3 \mathrm{~d}^2$ Hybridization
3 Paramagnetic
4 Diamagnetic
Explanation:
(C) : $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing occur because $\mathrm{CN}$ is SFL) There is one unpaired electron present. Hence, complex ion is paramagnetic.
274141
Which of the following outer octahedral complexes have same number of unpaired electrons? 1.$\left(\mathrm{MnCl}_{6}\right)^{3-}$ 3.$\left(\mathrm{CoF}_{6}\right)^{3-}$ 2.$\left(\mathrm{FeF}_{6}\right)^{3-}$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$
1 1 and 3
2 1 and 2
3 3 and 4
4 2 and 3
Explanation:
(A) : All four complexes given are outer orbital or high spin octahedral complexes. So, d-orbital electrons of central metal ions of the complexes remain unaffected by weak field ligands. 1.$\left[\mathrm{MnCl}_{6}\right]^{3-} \rightarrow \mathrm{Mn}^{3+} \Rightarrow \mathrm{d}^{4}, \mathrm{n}=4$ $[\because \mathrm{n}=$ Number of unpaired electrons $]$ 2.$\left[\mathrm{FeF}_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+} \Rightarrow \mathrm{d}^{5}, \mathrm{n}=5$ 3.$\left[\mathrm{CoF}_{6}\right]^{3-} \rightarrow \mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}, \mathrm{n}=4$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} \rightarrow \mathrm{Ni}^{2+} \Rightarrow \mathrm{d}^{8}, \mathrm{n}=2$ So, complexes 1 and 3 have same number of unpaired electrons $(\mathrm{n}=4)$.
AP EAMCET (Engg.) 17.09.2020
COORDINATION COMPOUNDS
274142
The ion that has $\mathbf{s p}^{3} \mathrm{~d}^{2}$ hybridization for the central atom,
274144
The complexes $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are, respectively
1 tetrahedral and square planar
2 tetrahedral and tetrahedral
3 square planar and square planar
4 square planar and tetrahedral.
Explanation:
(A) $\text { : } {\left[\mathrm{Ni}(\mathrm{CO})_4\right] }$ $x+4(0)=0$ $x=0$ Pairing of electron takes place because CO is SFL. Pairing of electron takes place because $\mathrm{CO}$ is $\mathrm{SFL}$. $\mathrm{Ni}=[\operatorname{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2 4 \mathrm{p}^0$ Hence, the complexes $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are tetrahedral and square planar respectively And, $\quad\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ $x+4(-1)=-2$ $x=+2$ Pairing of electron takes place because CN is SFL. $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
AMU-2019
COORDINATION COMPOUNDS
274147
Regarding $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{--}$complex ion which of the following statements is true?
1 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is paramagnetic.
2 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is diamagnetic
3 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is diamagnetic
4 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is paramagnetic.
Explanation:
(B) : $\left(\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $\quad \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 \text { (Pairing of electron occur because } \mathrm{CN}$ $\text { is SFL). }$
Tripura JEE-2019
COORDINATION COMPOUNDS
274148
Which of the given options are correct for $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ complex?
1 $\mathrm{d}^2 \mathrm{sp}^2$ Hybridization
2 $\mathrm{sp}^3 \mathrm{~d}^2$ Hybridization
3 Paramagnetic
4 Diamagnetic
Explanation:
(C) : $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing occur because $\mathrm{CN}$ is SFL) There is one unpaired electron present. Hence, complex ion is paramagnetic.
274141
Which of the following outer octahedral complexes have same number of unpaired electrons? 1.$\left(\mathrm{MnCl}_{6}\right)^{3-}$ 3.$\left(\mathrm{CoF}_{6}\right)^{3-}$ 2.$\left(\mathrm{FeF}_{6}\right)^{3-}$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$
1 1 and 3
2 1 and 2
3 3 and 4
4 2 and 3
Explanation:
(A) : All four complexes given are outer orbital or high spin octahedral complexes. So, d-orbital electrons of central metal ions of the complexes remain unaffected by weak field ligands. 1.$\left[\mathrm{MnCl}_{6}\right]^{3-} \rightarrow \mathrm{Mn}^{3+} \Rightarrow \mathrm{d}^{4}, \mathrm{n}=4$ $[\because \mathrm{n}=$ Number of unpaired electrons $]$ 2.$\left[\mathrm{FeF}_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+} \Rightarrow \mathrm{d}^{5}, \mathrm{n}=5$ 3.$\left[\mathrm{CoF}_{6}\right]^{3-} \rightarrow \mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}, \mathrm{n}=4$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} \rightarrow \mathrm{Ni}^{2+} \Rightarrow \mathrm{d}^{8}, \mathrm{n}=2$ So, complexes 1 and 3 have same number of unpaired electrons $(\mathrm{n}=4)$.
AP EAMCET (Engg.) 17.09.2020
COORDINATION COMPOUNDS
274142
The ion that has $\mathbf{s p}^{3} \mathrm{~d}^{2}$ hybridization for the central atom,
274144
The complexes $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are, respectively
1 tetrahedral and square planar
2 tetrahedral and tetrahedral
3 square planar and square planar
4 square planar and tetrahedral.
Explanation:
(A) $\text { : } {\left[\mathrm{Ni}(\mathrm{CO})_4\right] }$ $x+4(0)=0$ $x=0$ Pairing of electron takes place because CO is SFL. Pairing of electron takes place because $\mathrm{CO}$ is $\mathrm{SFL}$. $\mathrm{Ni}=[\operatorname{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2 4 \mathrm{p}^0$ Hence, the complexes $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are tetrahedral and square planar respectively And, $\quad\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ $x+4(-1)=-2$ $x=+2$ Pairing of electron takes place because CN is SFL. $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
AMU-2019
COORDINATION COMPOUNDS
274147
Regarding $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{--}$complex ion which of the following statements is true?
1 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is paramagnetic.
2 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is diamagnetic
3 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is diamagnetic
4 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is paramagnetic.
Explanation:
(B) : $\left(\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $\quad \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 \text { (Pairing of electron occur because } \mathrm{CN}$ $\text { is SFL). }$
Tripura JEE-2019
COORDINATION COMPOUNDS
274148
Which of the given options are correct for $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ complex?
1 $\mathrm{d}^2 \mathrm{sp}^2$ Hybridization
2 $\mathrm{sp}^3 \mathrm{~d}^2$ Hybridization
3 Paramagnetic
4 Diamagnetic
Explanation:
(C) : $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing occur because $\mathrm{CN}$ is SFL) There is one unpaired electron present. Hence, complex ion is paramagnetic.
274141
Which of the following outer octahedral complexes have same number of unpaired electrons? 1.$\left(\mathrm{MnCl}_{6}\right)^{3-}$ 3.$\left(\mathrm{CoF}_{6}\right)^{3-}$ 2.$\left(\mathrm{FeF}_{6}\right)^{3-}$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$
1 1 and 3
2 1 and 2
3 3 and 4
4 2 and 3
Explanation:
(A) : All four complexes given are outer orbital or high spin octahedral complexes. So, d-orbital electrons of central metal ions of the complexes remain unaffected by weak field ligands. 1.$\left[\mathrm{MnCl}_{6}\right]^{3-} \rightarrow \mathrm{Mn}^{3+} \Rightarrow \mathrm{d}^{4}, \mathrm{n}=4$ $[\because \mathrm{n}=$ Number of unpaired electrons $]$ 2.$\left[\mathrm{FeF}_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+} \Rightarrow \mathrm{d}^{5}, \mathrm{n}=5$ 3.$\left[\mathrm{CoF}_{6}\right]^{3-} \rightarrow \mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}, \mathrm{n}=4$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} \rightarrow \mathrm{Ni}^{2+} \Rightarrow \mathrm{d}^{8}, \mathrm{n}=2$ So, complexes 1 and 3 have same number of unpaired electrons $(\mathrm{n}=4)$.
AP EAMCET (Engg.) 17.09.2020
COORDINATION COMPOUNDS
274142
The ion that has $\mathbf{s p}^{3} \mathrm{~d}^{2}$ hybridization for the central atom,
274144
The complexes $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are, respectively
1 tetrahedral and square planar
2 tetrahedral and tetrahedral
3 square planar and square planar
4 square planar and tetrahedral.
Explanation:
(A) $\text { : } {\left[\mathrm{Ni}(\mathrm{CO})_4\right] }$ $x+4(0)=0$ $x=0$ Pairing of electron takes place because CO is SFL. Pairing of electron takes place because $\mathrm{CO}$ is $\mathrm{SFL}$. $\mathrm{Ni}=[\operatorname{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2 4 \mathrm{p}^0$ Hence, the complexes $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are tetrahedral and square planar respectively And, $\quad\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ $x+4(-1)=-2$ $x=+2$ Pairing of electron takes place because CN is SFL. $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
AMU-2019
COORDINATION COMPOUNDS
274147
Regarding $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{--}$complex ion which of the following statements is true?
1 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is paramagnetic.
2 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is diamagnetic
3 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is diamagnetic
4 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is paramagnetic.
Explanation:
(B) : $\left(\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $\quad \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 \text { (Pairing of electron occur because } \mathrm{CN}$ $\text { is SFL). }$
Tripura JEE-2019
COORDINATION COMPOUNDS
274148
Which of the given options are correct for $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ complex?
1 $\mathrm{d}^2 \mathrm{sp}^2$ Hybridization
2 $\mathrm{sp}^3 \mathrm{~d}^2$ Hybridization
3 Paramagnetic
4 Diamagnetic
Explanation:
(C) : $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing occur because $\mathrm{CN}$ is SFL) There is one unpaired electron present. Hence, complex ion is paramagnetic.
274141
Which of the following outer octahedral complexes have same number of unpaired electrons? 1.$\left(\mathrm{MnCl}_{6}\right)^{3-}$ 3.$\left(\mathrm{CoF}_{6}\right)^{3-}$ 2.$\left(\mathrm{FeF}_{6}\right)^{3-}$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$
1 1 and 3
2 1 and 2
3 3 and 4
4 2 and 3
Explanation:
(A) : All four complexes given are outer orbital or high spin octahedral complexes. So, d-orbital electrons of central metal ions of the complexes remain unaffected by weak field ligands. 1.$\left[\mathrm{MnCl}_{6}\right]^{3-} \rightarrow \mathrm{Mn}^{3+} \Rightarrow \mathrm{d}^{4}, \mathrm{n}=4$ $[\because \mathrm{n}=$ Number of unpaired electrons $]$ 2.$\left[\mathrm{FeF}_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+} \Rightarrow \mathrm{d}^{5}, \mathrm{n}=5$ 3.$\left[\mathrm{CoF}_{6}\right]^{3-} \rightarrow \mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}, \mathrm{n}=4$ 4.$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} \rightarrow \mathrm{Ni}^{2+} \Rightarrow \mathrm{d}^{8}, \mathrm{n}=2$ So, complexes 1 and 3 have same number of unpaired electrons $(\mathrm{n}=4)$.
AP EAMCET (Engg.) 17.09.2020
COORDINATION COMPOUNDS
274142
The ion that has $\mathbf{s p}^{3} \mathrm{~d}^{2}$ hybridization for the central atom,
274144
The complexes $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are, respectively
1 tetrahedral and square planar
2 tetrahedral and tetrahedral
3 square planar and square planar
4 square planar and tetrahedral.
Explanation:
(A) $\text { : } {\left[\mathrm{Ni}(\mathrm{CO})_4\right] }$ $x+4(0)=0$ $x=0$ Pairing of electron takes place because CO is SFL. Pairing of electron takes place because $\mathrm{CO}$ is $\mathrm{SFL}$. $\mathrm{Ni}=[\operatorname{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2 4 \mathrm{p}^0$ Hence, the complexes $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are tetrahedral and square planar respectively And, $\quad\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ $x+4(-1)=-2$ $x=+2$ Pairing of electron takes place because CN is SFL. $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
AMU-2019
COORDINATION COMPOUNDS
274147
Regarding $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{--}$complex ion which of the following statements is true?
1 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is paramagnetic.
2 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is diamagnetic
3 $\mathrm{Ni}^{+2}$ ion is $\mathrm{sp}^3$-hybridized and complex ion is diamagnetic
4 $\mathrm{Ni}^{+2}$ ion is $\mathrm{dsp}^2$-hybridized and complex ion is paramagnetic.
Explanation:
(B) : $\left(\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ Let, $x$ be the oxidation state of $\mathrm{Ni}$. $\quad \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 \text { (Pairing of electron occur because } \mathrm{CN}$ $\text { is SFL). }$
Tripura JEE-2019
COORDINATION COMPOUNDS
274148
Which of the given options are correct for $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ complex?
1 $\mathrm{d}^2 \mathrm{sp}^2$ Hybridization
2 $\mathrm{sp}^3 \mathrm{~d}^2$ Hybridization
3 Paramagnetic
4 Diamagnetic
Explanation:
(C) : $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing occur because $\mathrm{CN}$ is SFL) There is one unpaired electron present. Hence, complex ion is paramagnetic.
