(A) 0.35 A0 (B) 0.56 A0 (C) 1.98A0 (D) 10.48A0
The covalent
radius of B is calculated by the relation given by Stevenson & Schoemaker,
given as:
For a
diatomic Hetero molecule:
Bond
Length (lA-B) = rA + rB- 0.09(XA-XB)
Where XA=
Electronegativity of more electronegative atom
Where XB=
Electronegativity of less electronegative atom
Bond
Length=1.46 A0 = 0.99 A0 + rB -
0.09(3.5-2.5)
Hence rB
= 0.56 A0
Solved Questions:
Conditions for back bonding:
(1) Both of the atoms bonded with back Bonding are must be present in 2nd-2nd or 2nd-3rd period. 4rth period onward back bonding does not take place.
(2) One of
the atoms has lone pair (donor atom) and another (acceptor atom) have vacant
Orbital and direction of back Bonding depends upon vacant Orbital.
(3) The donor atom must have localized donatable electron pair and there should be inter electronic repulsion (smaller size). In general these are later half second period P - block elements
(F, O, N and C).
(4) The acceptor atom must have low energy empty orbital which
generally are np or nd orbitals. Small and similar size orbital’s favour overlap.
(5)
Back bonding is a weak pi bond thus only effective overlapping will be form
back bonding.
(6) Back bonding is found to be effective and considerable in following type of overlapping.
(i)
2p-2p
(ii) 2p-3p
(iii) 2p-3d
(7) The extent of overlapping order is [2p-2p> 2p-3d >2p-3p]
(8) dx2-y2 and dx2 Orbital’s does not participate in back Bonding.
Solution:
In case of trisilyl phosphine [P(SiH3)3] Phosphorous (P) being larger in atomic
size so it does not face much interelectronic repulsion and so, not much eager
to donate it’s loan pair electron and comfortable with it. So, it will not
donate to Si atom in this case. Thus, back bonding will not take place in trisilyl
phosphine [P(SiCH3)3].
But in the case of trisilyl amine [N(SiH3)3],
the nitrogen (N)
atom is
smaller in atomic size leading to high interelectronic repulsion, so it
wants ease by donating it’s loan pair electron to other atoms near to
it. And Si has vacant d-orbital, so needy for it. Since
both conditions are follow here, hence back
bonding will take place.
The probability of an electron which allows an electron to get close to the nucleus is known as penetration effect. It is known as the proximity of the electron in the orbital to the nucleus.
(1) The relative penetration power of sub shells
within same shell (same value of n) follow the order as:
s>p>d>f
We consider it for each shell and sub shell as the relative density of the electrons near the nucleus of atoms. It is clear that the ‘S’ electrons have greater probability of coming closer to the nucleus than the P, d or f electrons of the same principal energy shell.
(2) For different values of shell
(n) and sub shell (l), decreasing penetrating power of an electron follows as:
1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....
(3) In other words, ‘S’ electrons penetrate (more nearer) more towards the nucleus than the ‘P’ electrons and the penetrating power of the electrons in the given principal energy shell varies as S> P> d>f. Thus the ‘S’ electrons experience more attraction from the nucleus than the p d or f electrons of the same principal energy shell.
Therefore, greater energy required to remove out electron from‘s’ orbital than ‘p’, d and ‘f’ orbital. Thus the ionization potential for pulling out an ‘s’ electron is maximum and it decreases in pulling out a p , d or f electron of the same principal energy shell.
(4) Ionization potential or Ionization enthalpy of an atom is directly proportional to penetration power of orbitals.
Application of Penetration effect:
Example: Ionization energy of Boron is smaller than Beryllium even though effective nuclear charge is higher?
Solution: The
electronic configurations of Boron and Beryllium are (5B=1S2,2S2,2p1)
and (4Be =1S2,2S2).
In Boron the
outermost electron is present in the 2p orbital (low penetration power) and is
less strongly bound than the electron present in a 2S orbital of Beryllium(Have
more penetration power), which will has a higher Zeff. It is easier to ionize
the Boron atom.
Related Questions:
Metallic radius or Crystal radius:
(1) The term crystal radius is used to denote the
size of atoms in metal.
(2) Metal atoms are closely packed spheres in
metallic crystal. The metal atoms are supposed to touch one another in crystal.
(3) Metallic radius is defined as one half the distances
between the centres of the nuclei of two atoms in a metallic crystal.
(4) Metallic radius is determined by X-ray diffraction
method.
(5) Metallic radii are about 10 to 15 % higher
than the single bond covalent raddi of those elements. Thus single bond covalent radius is smaller than the metallic
radius due to the no overlapping of atomic orbital in metallic bond.
Vander Waal’s radius > Metallic radius > Covalent radius
(6) For the
simplicity the term atomic radius is used for covalent radius as well as
metallic radius depending on whether the element is a non-metal or metal. However,
the atomic radii of inert gases are expressed in the terms of Vander Waal’s
radii.
(7) Metallic
radius is inversely proportional to the metallic bond strength.
(8) More
metallic radius –loose crystal packing-less bond strength. (BCC)
(9) Less
metallic radius –Tight crystal packing (FCC) - high bond strength.(HCP)
(10) For
non-metal, atomic radius means covalent radius.
(11) For metal,
atomic radius means metallic radius.
(12) For inert gases, atomic radius means Vander Waal’s radius.
