# The D and F Block Elements

## Chemistry

### NCERT

1   Write down the electronic configuration of:$\\$ ( i ) $Cr^{3+}$$\\ ( iii ) Cu^+$$\\$ ( v )$Co^{2+}$$\\ ( vii ) Mn^{2+}$$\\$ ( ii ) $Pm ^{3 +}$$\\ ( iv ) Ce ^{4+}$$\\$ ( vi )$Lu^{2+}$$\\ ( viii ) Th^{4 +} ##### Solution : (i) Cr^{3+} : 1s^22s^22p^63s^23p^63d^3$$\\$ (ii)$Pm^{3+}:1s^22s^2p^63s^23p^63d^{10}4s^24P^64d^{10}5s^25p^64f^4$$\\ (iii)Cu^+:1s^22s^22p^63s^23P^63d^{10}$$\\$ (iv)$Ce^{4+}:1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^6$$\\ (v)Co^{2+}:1s^22s^22p^63s^23p^63d^7$$\\$ Or, $[Ar]^{18}3d^7$$\\ (vi)Lu^{2+}:1s622s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^1$$\\$ (vii) $Mn^{2+}:1s^22s^22p^63s^23p^63d^5$$\\ (viii)Th^{4+}:1s^22s^22p^63s^23p^63d^{10}4s^24p^64p^{10}4f^{14}5s^25p^65d^{10}6s^26p^6 2 Why are Mn^{2+} compounds more stable than Fe^{2+} towards oxidation to their +3 state? ##### Solution : Electronic configuration of Mn^{2+} is [ Ar ] 3 d ^6$$\\$ Electronic configuration of $Fe^{2 +}$ is $[ Ar ] 3 d^6$$\\ It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d^5 configuration. This is the reason Mn ^{2+} shows resistance to oxidation to Mn^{ 3+} .\\ Also, Fe^{2+} has 3d^6 configuration and by losing one electron, its configuration changes to a more stable 3d^5 configuration. Therefore, Fe^{2+} easily gets oxidized to Fe^{2+} oxidation state. 3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number? ##### Solution : The oxidation states displayed by the first half of the first row of transition metals are given in the table below.\\ It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.\\ Sc(+2)=3d^1$$\\$ $Ti(+2)=3d^2$$\\ V(+2)=3d^3$$\\$ $Cr(+2)=3d^4$$\\ Mn(+2)=3d^5$$\\$ $+2$oxidation state is attained by the loss of the two $4s$ electrons by these metals. Since the number of d electrons in $(+2)$ state also increases from $Ti(+2)$ to $Mn(+ 2)$, the stability of $+2$ state increases (as d-orbital is becoming more and more half-filled). $Mn (+2)$ has $3d^5$ electrons (that is half-filled d shell, which is highly stable).

4   To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.