Valence Electronic Configuration of M

To determine the number of electrons present in the valence shell d-orbital of M, we need to first find out the oxidation state of M in the given complexes.

Oxidation state of M in K2[MC14]J:
The complex K2[MC14]J is a binary metal halide, where M has a coordination number of 6 and is surrounded by four chloride ions and two J ions. Here, the overall charge on the complex is 0, and the charge on K and J ions are +1 and -1 respectively.

The formula of the complex can be written as:
K2[MC14]J = K+K+MCl4-Cl-Cl-Cl-J-J

Therefore, the charge on MC14 unit is -1. The formula weight of MC14 is (M + 4 × Cl) = MCl4, which means that the oxidation state of M is +4.

Oxidation state of M in Ca[M(CN)6]:
The complex Ca[M(CN)6] is a hexacyanometallate, where M has a coordination number of 6 and is surrounded by six cyanide ions. Here, the overall charge on the complex is -2, and the charge on Ca ion is +2.

The formula of the complex can be written as:
Ca[M(CN)6] = Ca2+M(CN)6-6CN-

Therefore, the charge on M(CN)6 unit is -2. The formula weight of M(CN)6 is (M + 6 × CN) = MCN6, which means that the oxidation state of M is +2.

Calculation of Number of Electrons in Valence d-Orbital

Now that we know the oxidation state of M in the given complexes, we can use the spin-only magnetic moment to calculate the number of electrons present in the valence d-orbital of M.

The spin-only magnetic moment (μ) can be calculated using the formula:
μ = √n(n+2)BM

where n is the number of unpaired electrons in the d-orbital and BM is the Bohr magneton.

For K2[MC14]J:
μ = 4.9 BM
n(n+2) = (4.9/BM)^2
n(n+2) = 24.01
n ≈ 5

For Ca[M(CN)6]:
μ = 2.83 BM
n(n+2) = (2.83/BM)^2
n(n+2) = 7.99
n ≈ 2

Therefore, the number of electrons present in the valence d-orbital of M in K2[MC14]J and Ca[M(CN)6] are 5 and 2 respectively.