Ncert Exercises & Solutions

Answer the following questions

1. Which element of the transition series has highest second ionisation enthalpy?

2. Which element of the first transition series as highest third ionisation enthalpy?

3. Which element of the first transition series as lowest third ionisation enthalpy?

4. Identify the metal and justify your answer.

(I) Carbonyl M ${(C O)}_{5}$

(II) $M O_{3} F$

NEETprep Answer: (a) (i) Cu, because the electronic configuration of Cu is

3 d^{10} 4 s^{1}

. So, second electron needs to be removed from completely filled d-orbital whichis very difficult.

(ii) Zinc, because of electronic configuration of Zn = 3

d^{10} 4 s^{2} a n d Z n^{2 +} = 3 d^{10}

which is fully filled and hence is very stable. Removal of third electron requires very high energy.

(iii) Zinc, because of it has completely filled 3d subshell and no unpaired electron is available for metallic bonding.

(b) (i) Carbonyl M

{(C O)}_{5} i s F e {(C O)}_{5}

According to EAN rule, the effective number of a metal in a metal carbonyl is equal to the atomic numberof nearest inert gas EAN is caiculated as
EAN =numberof electrons in metal +2 x (CO)
=atomic numberof nearestinert gas

(ii)

M O_{3} F i s M n O_{3} F

I n M O_{3} F

Let us assume that oxidation state of M is x

x + 3

\times

(-2) - + (-1) - 0

or, x = +74 i.e, M is in +7 oxidation state of +7

Hence, the given compound is

M n O_{3} F

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