D and F block

Question 1

why are zn, Cd, Hg not transition elements

Answer 1

bc they have full d10 config in their ground state as well as oxidation state

Question 2

general electronic config of outer orbitals of d block

Answer 2

(n-1)d1-10 ns1-2

Question 3

why is the E.C of Cr : 3d5 4s1 and not 3d4 4s2

Answer 3

very little energy decrease between n-1 d orbitals and ns orbitals. also half or completely filled are more stable

Question 4

On what ground do u say that sc(21) is a transition element not zinc (30)

Answer 4

on the basis of incompletely filled d orbitals. In case of scandium in its ground state (3d1) is regarded as a transition element . on the other hand zn has completely filled d orbital(3d10) in its ground state as well as in its oxidation state. Hence not regarded as transition element

Question 5

SIlver atom has completely fill d10 config in its ground state. how can u say it is a transition element?

Answer 5

silver can exhibit +2 oxidation state where in it will have incompletely filled d subshell

Question 6

Why are transition metals very hard? and have low volatility

Answer 6

bc of the greater number of unpaired e- from n-1d the interatomic bonding increases their hardness increases

Question 7

zinc (Zn), mercury (Hg) and cadmium (Cd) are not very hard

Answer 7

they do not possess unpaired d-electrons.

Question 8

The metals chromium (Cr), molybdenum (Mo) and tungsten (W) are hardest ?

Answer 8

because they have greater bo. of unpaired electrons and high enthalpies of atomization.

Question 9

Why the second and third series have greater atomisation enthalpy

Answer 9

because of stronger metal-metal bond

Question 10

why ions of same charge in a given series show progressive decrease in radius with increasing atomic no?

Answer 10

. the new electrons enter a d orbital each time the nuclear charge increases by unity.
. d orbitals have poor shielding effect
. hence, the net electrostatic attraction between nuclear charge and outermost electron increases and ionic radius decreases

Question 11

why is the atomic radii of second series and third series similar

Answer 11

It is due to Lanthanoid contraction. It arises due to poor shielding effect of d and f electron.

Question 12

Why does Hf(159) and Zr(160) exhibits similar radii and very similar chemical and physical properties much more than expected

Answer 12

because of lanthanoid contraction. Electrons present in f subshell have poor shielding due to which with the increasing atomic number or increasing effective nuclear charge size gets constricted

Question 13

why titanium to zinc increase in density is noted?

Answer 13

small and irregular decrease in metallic radii coupled with the relative increase in atomic mass

Question 14

Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series

Answer 14

as these have their electrons of outer most shell at greater distance from the nucleus, as compared to atoms of 3d transition metals.

Question 15

why ionisation enthalpies along 3d transition series show irregular trend?

Answer 15

the removal of electrons alters the relative energies of 4s and 3d orbitals.

Question 16

why does ionisation enthalpy along periods in 3d series increase slightly?

Answer 16

1. as we move along Sc to Zn nuclear charge increases.
2. but e- are added to inner subshell i.e. 3d
   3.the 3d shield 4s from nuclear charge somewhat effectively than outer shell e- can shield each other.
3. thus ionic radii decrease less rapidly and ionisation enthalpy increases only slightly

Question 17

why is increase in second ionisation enthalpy expected?

Answer 17

as the effective nuclear charge increases because one d electron does not shield another electron from the influence of nuclear charge because d orbitals differ in direction.

Question 18

where does the steady increase in second and third ionisation enthalpy increases

Answer 18

1. mn2+ - second enthalpy
2. fe3+ - third enthalpy

both have d5 config