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 $3$d-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.

Solution :

The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states $(+2$ to $+7)$. The stability of $+2$ oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show $+2$ oxidation state. Its electronic configuration is $4s^2$ $3d^1$ . It loses all the three electrons to form Sc $3+ . +3$ oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, $[Ar]. Ti (+ 4)$ and $V(+5)$ are very stable for the same reason. For $Mn, +2$ oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, $[Ar] 3d^6$$\\$

5   What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: and $3 d^3 , 3 d^5 ,3 d^8$ and $3 d^4$ ?

Solution :

$\\$$\text{Electronic configuration in ground state} \ \ \ \ \ \ \text{Stable oxidation state}$$\\$ $3d^3?\text{(vanadium)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2,+3,+4 \text{and} +5$$\\$ $3d^5\text{(chromium)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +3,+4,6+$$\\$ $3d^5\text{(manganese)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2,+4,+6,+7$$\\$ $3d^8\text{(cobalt)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2,+3$$\\$ $3d^4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{There is no $3rd^4$ configuration in ground state}$