Chemistry

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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the image

 

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

K19 = 1s22s22p63s23p64s1

For 4s electron

n = 4

l = 0

m = 0

s = 1 2

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

(a) [Cr (H2O)6]+3 → Cr+3 → t 2 g 3 e g °  

(b) [Fe (H2O)6]+3 → Fe3+ → t 2 g 3 e g 2  

(c) [Ni (H2O)6]+2 → Ni2+ → t 2 g 6 e g 2  

(d) [V (H2O)6]+3 → V3+ → t 2 g 2 e g °  

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

Polarisation power µ Charge S i z e for K+, polarising power is least and ionic character is maximum.

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a year ago

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alok kumar singh

Contributor-Level 10

Reducing character increases from NH3 to BiH3.

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

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

Ni2+ : 4s03d8 (No pairing with Cl–)

[Ni (CO)4] : 4s03d10 (diamagnetic)

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

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

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

BF3 : sp2 hybridised ->trigonal planar

 

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

XeF2 : Hybrid orbital = s bond + L.P

= 2 + 3

= 5 = sp3d

Shape ->linear

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