Class 12th

Since spin only magnetic moment is 4.90BM so number of unpaired electrons must be 4, so If the complex is octahedral, then it has to be high spin complex with configuration t? g? e²g¹ in that case
CFSE = 4* (-0.4Δ? ) + 2*0.6Δ? = -0.4Δ?
If the complex is tetrahedral then its electronic configuration will be e? ², t? g² and CFSE will be = 3* (-0.6Δ? ) + 3* (0.4Δ? ) = -0.6Δ?

|A|≠0
For (P): A≠I?
So, A = [1 0; 0 1] or [1; 0 1] or [1 0; 1]
or [1; 1 0]
So (P) is false.
A = [1 0; 1 0] or [1; 0 1] or [1 0; 1]
⇒ tr (A)=2
⇒ Q is true

Two points on the line (say) x/3 = y/2, z=1 are (0,0,1) and (3,2,1)
So dr's of the line is (3,2,0)
Line passing through (1,2,1), parallel to L and coplanar with given plane is r = i+2j+k + t (3i+2j), t∈R (-2,0,1) satisfies the line (for t=-1)
⇒ (-2,0,1) lies on given plane.
Answer of the question is (B)
We can check other options by finding equation of plane
Equation plane: |x-1, y-2, z-1; 1+2, 2-0, 1-1; 2+2, 1-0, 2-1| = 0
⇒ 2 (x-1)-3 (y-2)-5 (z-1)=0
⇒ 2x-3y-5z+9=0

Reaction path is SN2 because OH? is strong nucleophile, but hydroxyl ion will not attack on chiral centres and so there is retension of configuration.

P? V? =P? V? 48*10? ³*3³ = P? *12³. P? =750*10?

AgBr=1.88g. Br=0.8g. %Br= (0.8/1.6)*100=50%.

Kindly consider the following Image

Δng=-1/2. Kp=Kc (RT)? ¹/². Kc=Kp (RT)¹/².

-I, -M effect of NO? increase reactivity towards nucleophilic addition reaction with HCN.