Pick out the correct statement with respect to $[Mn(CN)_6]^{3-}$ :

Question

What is the chemical formula of glycerol?

Answer 1

The question asks about the hybridization and geometry of the complex ion $[Mn(CN)_6]^{3-}$. To determine the correct hybridization and geometry, follow these steps:

Determine the oxidation state of the central metal atom:
- Cyanide ($CN^-$) is a ligand with a charge of $-1$.
- Let the oxidation state of manganese ($Mn$) be $x$.
- The complex ion has a net charge of $-3$, so: x + 6(-1) = -3
- This simplifies to: x - 6 = -3
- Therefore, x = +3.
Write the electronic configuration of the central metal ion:
- The electronic configuration of neutral Mn is $[Ar] 3d^5 4s^2$.
- In the +3 oxidation state, Mn loses three electrons, giving $[Ar] 3d^4$.
Determine the type of hybridization:
- Cyanide ($CN^-$) is a strong field ligand that causes pairing of electrons in the 3d orbitals.
- The electronic configuration after pairing becomes $[Ar] 3d^{3} 4s^0 4p^0 4d^0$.
- To form an octahedral complex, six vacant hybrid orbitals are needed.
- By using the 3d, 4s, and two 4p orbitals, the resulting hybridization is $d^2sp^3$.
Confirm the geometry:
- With $d^2sp^3$ hybridization, the geometry of the complex is octahedral.

Based on the above analysis, the correct statement is: "It is $d^2sp^3$ hybridised and octahedral."

Explanation of other options:

$sp^3d^2$ hybridized and tetrahedral: This is incorrect because tetrahedral complexes are generally not formed with this hybridization and number of ligands.
$dsp^2$ hybridized and square planar: This is incorrect because square planar complexes typically involve $d^8$ configurations, not $d^4$.
$sp^3d^2$ hybridized and octahedral: This is incorrect because this describes the geometrical configuration of complexes with empty d orbitals in the higher principal quantum number shell.

Answer 2

The question asks about the hybridization and geometry of the complex ion $[Mn(CN)_6]^{3-}$. To determine the correct hybridization and geometry, follow these steps:

Determine the oxidation state of the central metal atom:
- Cyanide ($CN^-$) is a ligand with a charge of $-1$.
- Let the oxidation state of manganese ($Mn$) be $x$.
- The complex ion has a net charge of $-3$, so: x + 6(-1) = -3
- This simplifies to: x - 6 = -3
- Therefore, x = +3.
Write the electronic configuration of the central metal ion:
- The electronic configuration of neutral Mn is $[Ar] 3d^5 4s^2$.
- In the +3 oxidation state, Mn loses three electrons, giving $[Ar] 3d^4$.
Determine the type of hybridization:
- Cyanide ($CN^-$) is a strong field ligand that causes pairing of electrons in the 3d orbitals.
- The electronic configuration after pairing becomes $[Ar] 3d^{3} 4s^0 4p^0 4d^0$.
- To form an octahedral complex, six vacant hybrid orbitals are needed.
- By using the 3d, 4s, and two 4p orbitals, the resulting hybridization is $d^2sp^3$.
Confirm the geometry:
- With $d^2sp^3$ hybridization, the geometry of the complex is octahedral.

Based on the above analysis, the correct statement is: "It is $d^2sp^3$ hybridised and octahedral."

Explanation of other options:

$sp^3d^2$ hybridized and tetrahedral: This is incorrect because tetrahedral complexes are generally not formed with this hybridization and number of ligands.
$dsp^2$ hybridized and square planar: This is incorrect because square planar complexes typically involve $d^8$ configurations, not $d^4$.
$sp^3d^2$ hybridized and octahedral: This is incorrect because this describes the geometrical configuration of complexes with empty d orbitals in the higher principal quantum number shell.

Pick out the correct statement with respect to $[Mn(CN)_6]^{3-}$ :

The Correct Option is B

Solution and Explanation

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