274133
The hybridization and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$, respectively are :
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{3} \mathrm{~d}$ and paramagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and diamagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}$ Let, O.N. of $\mathrm{Mn}=\mathrm{x}$ $\therefore \mathrm{x}+(-1) \times 6=-4$ $\mathrm{x}-6=-4$ $\mathrm{x}=+2$ $\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ Since, co-ordination no. is 6 . So it is an octahedral complex and $\mathrm{CN}^{-}$is a strong field ligand so hybridization will be $\mathrm{d}^{2} \mathrm{sp}^{3}$ ie., inner orbital complex and low spin. one unpaired electron is present so, it is paramagnetic
JEE Main 25-02-2021
COORDINATION COMPOUNDS
274134
What is the hybridization present in the complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3} \mathrm{~d}^{2}$
3 $\mathrm{dsp}^{3}$
4 $\mathrm{sp}^{3} \mathrm{~d}$
Explanation:
(A) : Given complex : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ Oxidation state of metal $=+3$ Geometry $=$ octahedral Electronic configuration of $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}=\mathrm{d}^{2} \mathrm{sp}^{3}$ Hybridized (Octahedral and diamagnetic]
TS- EAMCET 09.08.2021
COORDINATION COMPOUNDS
274135
The hybridisation and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, respectively are
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{2} \mathrm{~d}^{2}$ and diamagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and paramagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$ $x+6(-1)=-4$ $x=+2$ $\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing of electron occur because $\mathrm{CN}$ is SFL). Ground state Excited state There is one unpaired electron. six d'sp' hybrid orbital Hence, $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ is paramagnetic And, $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ The same method also apply for $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ because it contain $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ electronic configuration which is similar to $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ Thus, both are paramagnetic with $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization.
JEE Main 2021
COORDINATION COMPOUNDS
274136
According to the valence bond theory the hybridisation of central metal atom is $\mathrm{dsp}^{2}$ for which one of the following compound?
(B) : The hybridization of central atom is $\mathrm{dsp}^{2}$ that means the complex is square planar in geometry. $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals. $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ $2(+1)+\mathrm{x}+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274133
The hybridization and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$, respectively are :
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{3} \mathrm{~d}$ and paramagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and diamagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}$ Let, O.N. of $\mathrm{Mn}=\mathrm{x}$ $\therefore \mathrm{x}+(-1) \times 6=-4$ $\mathrm{x}-6=-4$ $\mathrm{x}=+2$ $\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ Since, co-ordination no. is 6 . So it is an octahedral complex and $\mathrm{CN}^{-}$is a strong field ligand so hybridization will be $\mathrm{d}^{2} \mathrm{sp}^{3}$ ie., inner orbital complex and low spin. one unpaired electron is present so, it is paramagnetic
JEE Main 25-02-2021
COORDINATION COMPOUNDS
274134
What is the hybridization present in the complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3} \mathrm{~d}^{2}$
3 $\mathrm{dsp}^{3}$
4 $\mathrm{sp}^{3} \mathrm{~d}$
Explanation:
(A) : Given complex : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ Oxidation state of metal $=+3$ Geometry $=$ octahedral Electronic configuration of $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}=\mathrm{d}^{2} \mathrm{sp}^{3}$ Hybridized (Octahedral and diamagnetic]
TS- EAMCET 09.08.2021
COORDINATION COMPOUNDS
274135
The hybridisation and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, respectively are
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{2} \mathrm{~d}^{2}$ and diamagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and paramagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$ $x+6(-1)=-4$ $x=+2$ $\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing of electron occur because $\mathrm{CN}$ is SFL). Ground state Excited state There is one unpaired electron. six d'sp' hybrid orbital Hence, $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ is paramagnetic And, $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ The same method also apply for $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ because it contain $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ electronic configuration which is similar to $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ Thus, both are paramagnetic with $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization.
JEE Main 2021
COORDINATION COMPOUNDS
274136
According to the valence bond theory the hybridisation of central metal atom is $\mathrm{dsp}^{2}$ for which one of the following compound?
(B) : The hybridization of central atom is $\mathrm{dsp}^{2}$ that means the complex is square planar in geometry. $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals. $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ $2(+1)+\mathrm{x}+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
274133
The hybridization and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$, respectively are :
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{3} \mathrm{~d}$ and paramagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and diamagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}$ Let, O.N. of $\mathrm{Mn}=\mathrm{x}$ $\therefore \mathrm{x}+(-1) \times 6=-4$ $\mathrm{x}-6=-4$ $\mathrm{x}=+2$ $\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ Since, co-ordination no. is 6 . So it is an octahedral complex and $\mathrm{CN}^{-}$is a strong field ligand so hybridization will be $\mathrm{d}^{2} \mathrm{sp}^{3}$ ie., inner orbital complex and low spin. one unpaired electron is present so, it is paramagnetic
JEE Main 25-02-2021
COORDINATION COMPOUNDS
274134
What is the hybridization present in the complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3} \mathrm{~d}^{2}$
3 $\mathrm{dsp}^{3}$
4 $\mathrm{sp}^{3} \mathrm{~d}$
Explanation:
(A) : Given complex : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ Oxidation state of metal $=+3$ Geometry $=$ octahedral Electronic configuration of $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}=\mathrm{d}^{2} \mathrm{sp}^{3}$ Hybridized (Octahedral and diamagnetic]
TS- EAMCET 09.08.2021
COORDINATION COMPOUNDS
274135
The hybridisation and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, respectively are
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{2} \mathrm{~d}^{2}$ and diamagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and paramagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$ $x+6(-1)=-4$ $x=+2$ $\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing of electron occur because $\mathrm{CN}$ is SFL). Ground state Excited state There is one unpaired electron. six d'sp' hybrid orbital Hence, $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ is paramagnetic And, $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ The same method also apply for $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ because it contain $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ electronic configuration which is similar to $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ Thus, both are paramagnetic with $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization.
JEE Main 2021
COORDINATION COMPOUNDS
274136
According to the valence bond theory the hybridisation of central metal atom is $\mathrm{dsp}^{2}$ for which one of the following compound?
(B) : The hybridization of central atom is $\mathrm{dsp}^{2}$ that means the complex is square planar in geometry. $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals. $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ $2(+1)+\mathrm{x}+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
274133
The hybridization and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$, respectively are :
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{3} \mathrm{~d}$ and paramagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and diamagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}$ Let, O.N. of $\mathrm{Mn}=\mathrm{x}$ $\therefore \mathrm{x}+(-1) \times 6=-4$ $\mathrm{x}-6=-4$ $\mathrm{x}=+2$ $\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ Since, co-ordination no. is 6 . So it is an octahedral complex and $\mathrm{CN}^{-}$is a strong field ligand so hybridization will be $\mathrm{d}^{2} \mathrm{sp}^{3}$ ie., inner orbital complex and low spin. one unpaired electron is present so, it is paramagnetic
JEE Main 25-02-2021
COORDINATION COMPOUNDS
274134
What is the hybridization present in the complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3} \mathrm{~d}^{2}$
3 $\mathrm{dsp}^{3}$
4 $\mathrm{sp}^{3} \mathrm{~d}$
Explanation:
(A) : Given complex : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ Oxidation state of metal $=+3$ Geometry $=$ octahedral Electronic configuration of $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}=\mathrm{d}^{2} \mathrm{sp}^{3}$ Hybridized (Octahedral and diamagnetic]
TS- EAMCET 09.08.2021
COORDINATION COMPOUNDS
274135
The hybridisation and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, respectively are
1 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic
2 $\mathrm{sp}^{2} \mathrm{~d}^{2}$ and diamagnetic
3 $\mathrm{d}^{2} \mathrm{sp}^{3}$ and diamagnetic
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and paramagnetic
Explanation:
(A) : $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$ $x+6(-1)=-4$ $x=+2$ $\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ (Pairing of electron occur because $\mathrm{CN}$ is SFL). Ground state Excited state There is one unpaired electron. six d'sp' hybrid orbital Hence, $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ is paramagnetic And, $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ The same method also apply for $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ because it contain $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ electronic configuration which is similar to $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ Thus, both are paramagnetic with $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization.
JEE Main 2021
COORDINATION COMPOUNDS
274136
According to the valence bond theory the hybridisation of central metal atom is $\mathrm{dsp}^{2}$ for which one of the following compound?
(B) : The hybridization of central atom is $\mathrm{dsp}^{2}$ that means the complex is square planar in geometry. $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals. $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ $2(+1)+\mathrm{x}+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 4 \mathrm{p}^0$
one unpaired electron is present so, it is paramagnetic
