274047
The geometry and magnetic behaviour of the complex $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are
1 square planar geometry and diamagnetic
2 tetrahedral geometry and diamagnetic
3 square planar geometry and paramagnetic
4 tetrahedral geometry and paramagnetic.
Explanation:
(B) : The complex is $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$. ${\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]}$ $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\mathrm{CO}$ is a strong field ligand, pairing will take place. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\because$ Hybridisation $=\mathrm{sp}^{3}$ $\therefore$ Geometry $=$ Tetrahedral Four $\mathrm{sp}^{3}$ hybrid orbital Four $\mathrm{sp}^{3}$ hybrid orbital shows the diamagnetic character because no unpaired $\mathrm{e}^{-}$is present.
NEET 2018
COORDINATION COMPOUNDS
274048
Which shows the maximum magnetic moment?
1 $\mathrm{V}^{3+}$
2 $\mathrm{Cr}^{3+}$
3 $\mathrm{Fe}^{3+}$
4 $\mathrm{Co}^{3+}$
Explanation:
(C) : $\mathrm{V}^{3+}: \mathrm{d}^2$ Number of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}=2.84$ BIM. Number of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=\sqrt{15}=3.8$ B.M. $\mathrm{Fe}^{3+}: \mathrm{d}^5$ Number of unpaired electron $=5$ Magnetic moment $=\sqrt{5(5+2)}=\sqrt{35}=5.9$ B.M. Number of unpaired electron $=4$ Magnetic moment $=\sqrt{4(4+2)}=\sqrt{24}=$ 4.9B.M. Hence, $\mathrm{Fe}^{3+}$ maximum magnetic moment.
(B) : (a) $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{2+}$ $\mathrm{CN}^{-}$act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ act as diamagnetic substance. (b) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{3+}$ Number of unpaired electron $=1$ Hence, it act as paramagnetic (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \Rightarrow \mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\mathrm{CO}$ act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ (d) $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ or $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ $\mathrm{CO}^{3+}: \mathrm{d}^{6}$ Number of unpaired electron $=0$
HP CET-2018
COORDINATION COMPOUNDS
274050
What will be the geometry and magnetic moment of the complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ ?
1 Tetrahedral and 3.87 B.M.
2 Tetrahedral and 2.82 B.M.
3 Square planar and 2.82 B.M.
4 Square planar and 4.89 B.M.
Explanation:
(B) : $\mathrm{Cl}^{-}$act as weak field ligand, and pairing is not possible. $\left[\mathrm{NiCl}_{4}\right]^{2-} \Rightarrow \mathrm{Ni}^{2+} ; \mathrm{d}^{8}$ $\mathrm{Cl}^{-}$occupies $\mathrm{sp}^{3}$ orbitals. Hybridization $=\mathrm{sp}^{3}$ Geometry $=$ tetrahedral No. of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}$ $=2.82$ B.M.
274047
The geometry and magnetic behaviour of the complex $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are
1 square planar geometry and diamagnetic
2 tetrahedral geometry and diamagnetic
3 square planar geometry and paramagnetic
4 tetrahedral geometry and paramagnetic.
Explanation:
(B) : The complex is $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$. ${\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]}$ $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\mathrm{CO}$ is a strong field ligand, pairing will take place. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\because$ Hybridisation $=\mathrm{sp}^{3}$ $\therefore$ Geometry $=$ Tetrahedral Four $\mathrm{sp}^{3}$ hybrid orbital Four $\mathrm{sp}^{3}$ hybrid orbital shows the diamagnetic character because no unpaired $\mathrm{e}^{-}$is present.
NEET 2018
COORDINATION COMPOUNDS
274048
Which shows the maximum magnetic moment?
1 $\mathrm{V}^{3+}$
2 $\mathrm{Cr}^{3+}$
3 $\mathrm{Fe}^{3+}$
4 $\mathrm{Co}^{3+}$
Explanation:
(C) : $\mathrm{V}^{3+}: \mathrm{d}^2$ Number of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}=2.84$ BIM. Number of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=\sqrt{15}=3.8$ B.M. $\mathrm{Fe}^{3+}: \mathrm{d}^5$ Number of unpaired electron $=5$ Magnetic moment $=\sqrt{5(5+2)}=\sqrt{35}=5.9$ B.M. Number of unpaired electron $=4$ Magnetic moment $=\sqrt{4(4+2)}=\sqrt{24}=$ 4.9B.M. Hence, $\mathrm{Fe}^{3+}$ maximum magnetic moment.
(B) : (a) $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{2+}$ $\mathrm{CN}^{-}$act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ act as diamagnetic substance. (b) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{3+}$ Number of unpaired electron $=1$ Hence, it act as paramagnetic (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \Rightarrow \mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\mathrm{CO}$ act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ (d) $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ or $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ $\mathrm{CO}^{3+}: \mathrm{d}^{6}$ Number of unpaired electron $=0$
HP CET-2018
COORDINATION COMPOUNDS
274050
What will be the geometry and magnetic moment of the complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ ?
1 Tetrahedral and 3.87 B.M.
2 Tetrahedral and 2.82 B.M.
3 Square planar and 2.82 B.M.
4 Square planar and 4.89 B.M.
Explanation:
(B) : $\mathrm{Cl}^{-}$act as weak field ligand, and pairing is not possible. $\left[\mathrm{NiCl}_{4}\right]^{2-} \Rightarrow \mathrm{Ni}^{2+} ; \mathrm{d}^{8}$ $\mathrm{Cl}^{-}$occupies $\mathrm{sp}^{3}$ orbitals. Hybridization $=\mathrm{sp}^{3}$ Geometry $=$ tetrahedral No. of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}$ $=2.82$ B.M.
274047
The geometry and magnetic behaviour of the complex $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are
1 square planar geometry and diamagnetic
2 tetrahedral geometry and diamagnetic
3 square planar geometry and paramagnetic
4 tetrahedral geometry and paramagnetic.
Explanation:
(B) : The complex is $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$. ${\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]}$ $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\mathrm{CO}$ is a strong field ligand, pairing will take place. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\because$ Hybridisation $=\mathrm{sp}^{3}$ $\therefore$ Geometry $=$ Tetrahedral Four $\mathrm{sp}^{3}$ hybrid orbital Four $\mathrm{sp}^{3}$ hybrid orbital shows the diamagnetic character because no unpaired $\mathrm{e}^{-}$is present.
NEET 2018
COORDINATION COMPOUNDS
274048
Which shows the maximum magnetic moment?
1 $\mathrm{V}^{3+}$
2 $\mathrm{Cr}^{3+}$
3 $\mathrm{Fe}^{3+}$
4 $\mathrm{Co}^{3+}$
Explanation:
(C) : $\mathrm{V}^{3+}: \mathrm{d}^2$ Number of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}=2.84$ BIM. Number of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=\sqrt{15}=3.8$ B.M. $\mathrm{Fe}^{3+}: \mathrm{d}^5$ Number of unpaired electron $=5$ Magnetic moment $=\sqrt{5(5+2)}=\sqrt{35}=5.9$ B.M. Number of unpaired electron $=4$ Magnetic moment $=\sqrt{4(4+2)}=\sqrt{24}=$ 4.9B.M. Hence, $\mathrm{Fe}^{3+}$ maximum magnetic moment.
(B) : (a) $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{2+}$ $\mathrm{CN}^{-}$act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ act as diamagnetic substance. (b) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{3+}$ Number of unpaired electron $=1$ Hence, it act as paramagnetic (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \Rightarrow \mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\mathrm{CO}$ act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ (d) $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ or $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ $\mathrm{CO}^{3+}: \mathrm{d}^{6}$ Number of unpaired electron $=0$
HP CET-2018
COORDINATION COMPOUNDS
274050
What will be the geometry and magnetic moment of the complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ ?
1 Tetrahedral and 3.87 B.M.
2 Tetrahedral and 2.82 B.M.
3 Square planar and 2.82 B.M.
4 Square planar and 4.89 B.M.
Explanation:
(B) : $\mathrm{Cl}^{-}$act as weak field ligand, and pairing is not possible. $\left[\mathrm{NiCl}_{4}\right]^{2-} \Rightarrow \mathrm{Ni}^{2+} ; \mathrm{d}^{8}$ $\mathrm{Cl}^{-}$occupies $\mathrm{sp}^{3}$ orbitals. Hybridization $=\mathrm{sp}^{3}$ Geometry $=$ tetrahedral No. of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}$ $=2.82$ B.M.
274047
The geometry and magnetic behaviour of the complex $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are
1 square planar geometry and diamagnetic
2 tetrahedral geometry and diamagnetic
3 square planar geometry and paramagnetic
4 tetrahedral geometry and paramagnetic.
Explanation:
(B) : The complex is $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$. ${\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]}$ $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\mathrm{CO}$ is a strong field ligand, pairing will take place. $\mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\because$ Hybridisation $=\mathrm{sp}^{3}$ $\therefore$ Geometry $=$ Tetrahedral Four $\mathrm{sp}^{3}$ hybrid orbital Four $\mathrm{sp}^{3}$ hybrid orbital shows the diamagnetic character because no unpaired $\mathrm{e}^{-}$is present.
NEET 2018
COORDINATION COMPOUNDS
274048
Which shows the maximum magnetic moment?
1 $\mathrm{V}^{3+}$
2 $\mathrm{Cr}^{3+}$
3 $\mathrm{Fe}^{3+}$
4 $\mathrm{Co}^{3+}$
Explanation:
(C) : $\mathrm{V}^{3+}: \mathrm{d}^2$ Number of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}=2.84$ BIM. Number of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=\sqrt{15}=3.8$ B.M. $\mathrm{Fe}^{3+}: \mathrm{d}^5$ Number of unpaired electron $=5$ Magnetic moment $=\sqrt{5(5+2)}=\sqrt{35}=5.9$ B.M. Number of unpaired electron $=4$ Magnetic moment $=\sqrt{4(4+2)}=\sqrt{24}=$ 4.9B.M. Hence, $\mathrm{Fe}^{3+}$ maximum magnetic moment.
(B) : (a) $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{2+}$ $\mathrm{CN}^{-}$act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ act as diamagnetic substance. (b) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow \mathrm{Fe}^{3+}$ Number of unpaired electron $=1$ Hence, it act as paramagnetic (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \Rightarrow \mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $\mathrm{CO}$ act as strong field ligand and do pairing of delectron. Number of unpaired electron $=0$ (d) $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ or $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ $\mathrm{CO}^{3+}: \mathrm{d}^{6}$ Number of unpaired electron $=0$
HP CET-2018
COORDINATION COMPOUNDS
274050
What will be the geometry and magnetic moment of the complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ ?
1 Tetrahedral and 3.87 B.M.
2 Tetrahedral and 2.82 B.M.
3 Square planar and 2.82 B.M.
4 Square planar and 4.89 B.M.
Explanation:
(B) : $\mathrm{Cl}^{-}$act as weak field ligand, and pairing is not possible. $\left[\mathrm{NiCl}_{4}\right]^{2-} \Rightarrow \mathrm{Ni}^{2+} ; \mathrm{d}^{8}$ $\mathrm{Cl}^{-}$occupies $\mathrm{sp}^{3}$ orbitals. Hybridization $=\mathrm{sp}^{3}$ Geometry $=$ tetrahedral No. of unpaired electron $=2$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}$ $=2.82$ B.M.
