Number of complexes from the following with eneven number of unpaired "d" electrons is ________.

[V(H₂O)₆]³⁺, [Cr(H₂O)₆] ²⁺, [Fe(H₂O)₆]³⁺ , [Ni(H₂O)₆]³⁺ , [Cu(H₂O)₆]²⁺

(Given atomic numbers : V = 23, Cr = 24, Fe = 26, Ni = 28, Cu = 29]

The correct answer is 2 (Option 1).

To find the number of complexes with an even number of unpaired d electrons (noting the typo "eneven" in the question which refers to "even"), we determine the oxidation state, electronic configuration, and number of unpaired electrons (ⁿ) for the central metal ion in each complex. Since H₂O is a weak field ligand, it does not cause pairing of electrons (resulting in high-spin complexes).

1. [V(H₂O)₆]³⁺
The atomic number of V is 23. The configuration of neutral V is [Ar] 3d³ 4s².
In [V(H₂O)₆]³⁺, the vanadium ion is in the +3 oxidation state:
V³⁺ = [Ar] 3d²
The distribution in the t_2g and e_g orbitals is:
$t_{\text{2g}}{}^2{e}_{\text{g}}{}^0$
The number of unpaired electrons is:
$n=2$ (Even)

2. [Cr(H₂O)₆]²⁺
The atomic number of Cr is 24. The configuration of neutral Cr is [Ar] 3d⁵ 4s¹.
In [Cr(H₂O)₆]²⁺, the chromium ion is in the +2 oxidation state:
Cr²⁺ = [Ar] 3d⁴
Since H₂O is a weak field ligand, the distribution is:
$t_{\text{2g}}{}^3{e}_{\text{g}}{}^1$
The number of unpaired electrons is:
$n=4$ (Even)

3. [Fe(H₂O)₆]³⁺
The atomic number of Fe is 26. The configuration of neutral Fe is [Ar] 3d⁶ 4s².
In [Fe(H₂O)₆]³⁺, the iron ion is in the +3 oxidation state:
Fe³⁺ = [Ar] 3d⁵
With H₂O as a weak field ligand, the distribution is:
$t_{\text{2g}}{}^3{e}_{\text{g}}{}^2$
The number of unpaired electrons is:
$n=5$ (Odd)

4. [Ni(H₂O)₆]³⁺
The atomic number of Ni is 28. The configuration of neutral Ni is [Ar] 3d⁸ 4s².
In [Ni(H₂O)₆]³⁺, the nickel ion is in the +3 oxidation state:
Ni³⁺ = [Ar] 3d⁷
With H₂O as a weak field ligand, the distribution is:
$t_{\text{2g}}{}^5{e}_{\text{g}}{}^2$
The number of unpaired electrons is:
$n=3$ (Odd)

5. [Cu(H₂O)₆]²⁺
The atomic number of Cu is 29. The configuration of neutral Cu is [Ar] 3d¹⁰ 4s¹.
In [Cu(H₂O)₆]²⁺, the copper ion is in the +2 oxidation state:
Cu²⁺ = [Ar] 3d⁹
The distribution is:
$t_{\text{2g}}{}^6{e}_{\text{g}}{}^3$
The number of unpaired electrons is:
$n=1$ (Odd)

Comparing the values, the complexes with an even number of unpaired electrons are [V(H₂O)₆]³⁺ (n = 2) and [Cr(H₂O)₆]²⁺ (n = 4).

Thus, the number of complexes with an even number of unpaired d electrons is 2.