D & F Block Elements MCQs for NEET — Chemistry Questions with Answers

The highest principal quantum number 'n' is 4, so the element is in Period 4. The configuration $ns^2np^4$ indicates a p-block element. For p-block elements, the group number is $10 + ext{number of s-electrons} + ext{number of p-electrons}$. So, Group = $10 + 2 + 4 = 16$. This element is Selenium (Se, Z=34).

Elements in the same vertical column or group have similar valence shell electronic configurations, the same number of electrons in the outer orbitals, and similar properties. This similarity in 'outermost orbital electrons' dictates their chemical reactivity and bonding patterns.

The 4f- and 5f-inner transition series of elements are placed separately in the Periodic Table to maintain its structure and to preserve the principle of classification by keeping elements with similar properties in a single column. Integrating them into the main body would make the periodic table unduly long and difficult to manage.

D-block elements are characterized by the filling of d-orbitals. The general configuration is $(n-1)d^{1-10}ns^{1-2}$. Option C, $[Ar]3d^74s^2$, shows the last electron entering a d-orbital ($3d$), indicating it's a d-block element (Cobalt, Co, Z=27). Options A and D are p-block elements, and Option B is an s-block element.

For $n=6$, the configuration becomes $4f^75d^16s^2$. The presence of filling f-orbitals (specifically $4f$) indicates it is an f-block element. Elements with $4f$ filling are called Lanthanoids. This configuration corresponds to Gadolinium (Gd, Z=64).

The text states: 'There are greater similarities in the properties of the transition elements of a horizontal row in contrast to the non-transition elements. However, some group similarities also exist.' This highlights a characteristic feature of transition metals where horizontal similarities are often more pronounced than vertical ones due to the gradual filling of inner d-orbitals.

For the 5th Period (n=5), the orbitals filled are $5s$, $4d$, and $5p$. The order of filling is $5s < 4d < 5p$. $5s$ can hold 2 electrons, $4d$ can hold 10 electrons, and $5p$ can hold 6 electrons. Therefore, the total number of electrons (and thus elements) is $2 + 10 + 6 = 18$. The NCERT text states: 'When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electrons that can be accommodated is 18; and therefore 18 elements are there in the 5th period.'

The context clearly states: 'Now according to IUPAC, transition metals are defined as metals which have incomplete d subshell either in neutral atom or in their ions.' The previous definition was based on 'transitional chemical properties' but the IUPAC definition focuses on the presence of an incompletely filled d-subshell.

The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7. (NCERT, Chapter: The d - and f - Block Elements, Page 97)

The lesser number of oxidation states at the extreme ends stems from either too few electrons to lose or share (Sc, Ti). Thus, early in the series scandium(II) is virtually unknown. (NCERT, Chapter: The d - and f - Block Elements, Page 97)