Δx.Δp ≧ h / 4π

So by the equation given in the question, the uncertainity value will be the least.

Maximum number of electrons can be accommodated in a shell of principal quantum number n is 2n². Since each orbital can accommodate only 2 electrons, maximum number of orbitals =2n²/2= n²

Total no. of electrons in F- subshell = 14 (when we include both spin +1/2  & - 1/2)

But maximum no. of electrons = 7 if only one spin either - 1/2 or +1/2.

Answer : The electronic configuration of Cu is,

1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰

The reason for abnormal electronic configuration of Cu is because of the half-filled and fully-filled configuration are the more stable configuration. In this configuration, the d-subshell is fully-filled and s-subshell is half-filled which makes the Cu more stable.

Now, for getting a positive charge, it has to lose 1 electron which will be removed from the outermost 4s orbital and hence all the remaining electrons will be paired leading to 0 unpaired electrons.

Two electrons in K-shell will differ in spin quantum number.

They will have same values for the principal quantum number, azimuthal quantum number, and magnetic quantum number.

For first electron, n = 1,l = 0,m = 0,s = +1 ​/ 2

For second electron, n = 1,l = 0,m = 0,s = −1 / 2

Total no of angular nodes = l (azimuthal quantum no), so l=2 in d orbital, therefore number of nodes is 2 and it is a fact that opposite lobes have same signs and adjacent ones have opposite signs. there are 4 lobes in d orbital.

Boiling Point is directly proportional to VanderWaals forces of attraction. Down the group, this forces increases. so force of attraction in CHF₃ is weaker than CHL₃, that's why boiling point of CHF₃ is lowest.

SiO₂ is a covalent 'network' solid. So the energy required to break bonds will be higher. Therefore the melting point will be higher.

Due to intramolecular H-bonding  in o-nitrophenol it exists as a monomer while due to intermolecular H-bonding in p-nitrophenol it exists as an associate molecule. As a result, vapour pressure of a o-nitrophenol is higher than that of a p-nitrophenol.

When the number of hybrid orbitals, H  is 4, the hybridization is sp³ hybridization

H = 1 / 2[V + M - C + A]

where, V = number of monovalent atoms

C = total positive charge

A = negative charge

a) For CH₄

H = 1 / 2[4 + 4 - 0 + 0] = 4

sp³ 

b) For SF₄,

H = 1 / 2 [6 + 4 - 0 + 0] = 5

sp³d

c) For BF^-₄,

H = 1 / 2 [3 + 4 - 0 + 1] = 4, thus sp³

d) For NH⁺₄

H = 1 / 2[5 + 4 - 1 + 0] = 4 thus sp³

Thus, only in SF₄ the central atom does not have sp³ hybridisation.