274361
What is the oxidation number of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ ?
1 +3
2 zero
3 +2
4 +5
Explanation:
(B) : Let the oxidation state of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ is $\mathrm{x}$. $\mathrm{Fe}(\mathrm{CO})_{5}$ $\mathrm{x}+0 \times 5=0$ $\therefore \quad \mathrm{x}=0$ In metal carbonyls, oxidation state of metal is always zero.
UP CPMT-2009
COORDINATION COMPOUNDS
274332
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:
1 $\mathrm{Ni}^{2+}$
2 $\mathrm{Fe}^{2+}$
3 $\mathrm{Co}^{2+}$
4 $\mathrm{Mn}^{2+}$
Explanation:
(C) : The compound is octahedral type of complex. Here, d-orbital split into two energy level i.e. $t_{2 g}$ and eg. Hence, $\mathrm{Co}^{2+}$ has the difference in oxidation state in low spin and spin, is +2
BITSAT-2021
COORDINATION COMPOUNDS
274337
Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is
1 +5
2 +1
3 +6
4 0
Explanation:
(D) : Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is 0 . Here $\mathrm{CO}$ is neutral ligand. $\mathrm{Mn}(\mathrm{CO})_{5}$ $\Rightarrow \quad \mathrm{x}+5(0)=0$ $\Rightarrow \quad \mathrm{x}=0$
JCECE - 2018
COORDINATION COMPOUNDS
274340
What should be possible d-orbital energy levels of $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}-$ ?
(C) : $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ forms the square planar geometry with CN ligand. The d-orbital energy levels illustrated below- Hence, the increasing order of energy can be written as $\mathrm{d}_{\mathrm{xz}}=\mathrm{d}_{\mathrm{yz}}<\mathrm{d}_{\mathrm{z}^{2}}<\mathrm{d}_{\mathrm{xy}}<\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}$
274361
What is the oxidation number of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ ?
1 +3
2 zero
3 +2
4 +5
Explanation:
(B) : Let the oxidation state of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ is $\mathrm{x}$. $\mathrm{Fe}(\mathrm{CO})_{5}$ $\mathrm{x}+0 \times 5=0$ $\therefore \quad \mathrm{x}=0$ In metal carbonyls, oxidation state of metal is always zero.
UP CPMT-2009
COORDINATION COMPOUNDS
274332
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:
1 $\mathrm{Ni}^{2+}$
2 $\mathrm{Fe}^{2+}$
3 $\mathrm{Co}^{2+}$
4 $\mathrm{Mn}^{2+}$
Explanation:
(C) : The compound is octahedral type of complex. Here, d-orbital split into two energy level i.e. $t_{2 g}$ and eg. Hence, $\mathrm{Co}^{2+}$ has the difference in oxidation state in low spin and spin, is +2
BITSAT-2021
COORDINATION COMPOUNDS
274337
Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is
1 +5
2 +1
3 +6
4 0
Explanation:
(D) : Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is 0 . Here $\mathrm{CO}$ is neutral ligand. $\mathrm{Mn}(\mathrm{CO})_{5}$ $\Rightarrow \quad \mathrm{x}+5(0)=0$ $\Rightarrow \quad \mathrm{x}=0$
JCECE - 2018
COORDINATION COMPOUNDS
274340
What should be possible d-orbital energy levels of $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}-$ ?
(C) : $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ forms the square planar geometry with CN ligand. The d-orbital energy levels illustrated below- Hence, the increasing order of energy can be written as $\mathrm{d}_{\mathrm{xz}}=\mathrm{d}_{\mathrm{yz}}<\mathrm{d}_{\mathrm{z}^{2}}<\mathrm{d}_{\mathrm{xy}}<\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274361
What is the oxidation number of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ ?
1 +3
2 zero
3 +2
4 +5
Explanation:
(B) : Let the oxidation state of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ is $\mathrm{x}$. $\mathrm{Fe}(\mathrm{CO})_{5}$ $\mathrm{x}+0 \times 5=0$ $\therefore \quad \mathrm{x}=0$ In metal carbonyls, oxidation state of metal is always zero.
UP CPMT-2009
COORDINATION COMPOUNDS
274332
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:
1 $\mathrm{Ni}^{2+}$
2 $\mathrm{Fe}^{2+}$
3 $\mathrm{Co}^{2+}$
4 $\mathrm{Mn}^{2+}$
Explanation:
(C) : The compound is octahedral type of complex. Here, d-orbital split into two energy level i.e. $t_{2 g}$ and eg. Hence, $\mathrm{Co}^{2+}$ has the difference in oxidation state in low spin and spin, is +2
BITSAT-2021
COORDINATION COMPOUNDS
274337
Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is
1 +5
2 +1
3 +6
4 0
Explanation:
(D) : Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is 0 . Here $\mathrm{CO}$ is neutral ligand. $\mathrm{Mn}(\mathrm{CO})_{5}$ $\Rightarrow \quad \mathrm{x}+5(0)=0$ $\Rightarrow \quad \mathrm{x}=0$
JCECE - 2018
COORDINATION COMPOUNDS
274340
What should be possible d-orbital energy levels of $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}-$ ?
(C) : $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ forms the square planar geometry with CN ligand. The d-orbital energy levels illustrated below- Hence, the increasing order of energy can be written as $\mathrm{d}_{\mathrm{xz}}=\mathrm{d}_{\mathrm{yz}}<\mathrm{d}_{\mathrm{z}^{2}}<\mathrm{d}_{\mathrm{xy}}<\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}$
274361
What is the oxidation number of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ ?
1 +3
2 zero
3 +2
4 +5
Explanation:
(B) : Let the oxidation state of $\mathrm{Fe}$ in $\mathrm{Fe}(\mathrm{CO})_{5}$ is $\mathrm{x}$. $\mathrm{Fe}(\mathrm{CO})_{5}$ $\mathrm{x}+0 \times 5=0$ $\therefore \quad \mathrm{x}=0$ In metal carbonyls, oxidation state of metal is always zero.
UP CPMT-2009
COORDINATION COMPOUNDS
274332
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:
1 $\mathrm{Ni}^{2+}$
2 $\mathrm{Fe}^{2+}$
3 $\mathrm{Co}^{2+}$
4 $\mathrm{Mn}^{2+}$
Explanation:
(C) : The compound is octahedral type of complex. Here, d-orbital split into two energy level i.e. $t_{2 g}$ and eg. Hence, $\mathrm{Co}^{2+}$ has the difference in oxidation state in low spin and spin, is +2
BITSAT-2021
COORDINATION COMPOUNDS
274337
Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is
1 +5
2 +1
3 +6
4 0
Explanation:
(D) : Oxidation state of manganese in $\mathrm{Mn}(\mathrm{CO})_{5}$ is 0 . Here $\mathrm{CO}$ is neutral ligand. $\mathrm{Mn}(\mathrm{CO})_{5}$ $\Rightarrow \quad \mathrm{x}+5(0)=0$ $\Rightarrow \quad \mathrm{x}=0$
JCECE - 2018
COORDINATION COMPOUNDS
274340
What should be possible d-orbital energy levels of $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}-$ ?
(C) : $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ forms the square planar geometry with CN ligand. The d-orbital energy levels illustrated below- Hence, the increasing order of energy can be written as $\mathrm{d}_{\mathrm{xz}}=\mathrm{d}_{\mathrm{yz}}<\mathrm{d}_{\mathrm{z}^{2}}<\mathrm{d}_{\mathrm{xy}}<\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}$
