Chemistry

K₁₉ = 1s²2s²2p⁶3s²3p⁶4s¹

For 4s electron

n = 4

l = 0

m = 0

$s=\frac{1}{2}$

(a) [Cr (H₂O)₆]⁺³ → Cr⁺³ → $t_{2g}^{3}eg°$  

(b) [Fe (H₂O)₆]⁺³ → Fe³⁺ → $t_{2g}^{3}e{g}^2$  

(c) [Ni (H₂O)₆]⁺² → Ni²⁺ → $t_{2g}^{6}e{g}^2$  

(d) [V (H₂O)₆]⁺³ → V³⁺ → $t_{2g}^{2}eg°$  

Polarisation power µ $\frac{Charge}{Size}$ for K⁺, polarising power is least and ionic character is maximum.

Reducing character increases from NH₃ to BiH₃.

Group 15 oxides of type E₂O₃ and E₂O₅ are not always basic.

Ni²⁺ : 4s⁰3d⁸ (No pairing with Cl^–)

[Ni (CO)₄] : 4s⁰3d¹⁰ (diamagnetic)

(2)  Cl is in its the highest oxidation state (+7). It cannot be further oxidised

BF₃ : sp² hybridised ->trigonal planar

XeF₂ : Hybrid orbital = s bond + L.P

= 2 + 3

= 5 = sp3d

Shape ->linear

Kindly go through the solution

Z=109 : Unnilennium->Meitnerium