Chemistry MCQs for NEET — Practice Questions with Answers

The NCERT text description of Reimer-Tiemann reaction says: 'The intermediate substituted benzal chloride is hydrolysed in the presence of alkali to produce salicylaldehyde.'

The context mentions, 'Phenoxide ion generated by treating phenol with sodium hydroxide is even more reactive than phenol towards electrophilic aromatic substitution.' This enhanced reactivity is due to the complete negative charge on oxygen in the phenoxide ion, which more effectively delocalizes electrons into the aromatic ring, making it more nucleophilic and reactive towards electrophiles than phenol where the oxygen has only partial negative charge.

The NCERT states: 'Phenol is converted to benzene on heating with zinc dust.'

The NCERT excerpt states: 'Oxidation of phenol with chromic acid produces a conjugated diketone known as benzoquinone.'

According to IUPAC, transition metals are defined as metals which have incomplete d subshell either in neutral atom or in their ions. Zinc (Z=30) has a full $3d^{10}$ configuration in its ground state as well as in its common oxidation states, and therefore is not regarded as a transition metal. Scandium (Z=21) has $3d^1$, Chromium (Z=24) has $3d^5$, and Copper (Z=29) has $3d^{10}$ (with $4s^1$), making their d-subshells incompletely filled in their elemental or common ionic states.

For $n=4$, the outer electronic configuration is $3d^24s^2$. The principal quantum number 'n' represents the period, so the element is in Period 4. The sum of electrons in the $(n-1)d$ and $ns$ orbitals gives the group number for d-block elements. Here, $2 (d) + 2 (s) = 4$. Thus, it is in Group 4. This corresponds to Titanium (Ti, Z=22).

The electronic configurations of Cr and Cu show exceptions due to the extra stability of half-filled ($d^5$) and completely filled ($d^{10}$) d-orbitals. Cr (Z=24) has the configuration $3d^54s^1$ instead of $3d^44s^2$. Cu (Z=29) has the configuration $3d^{10}4s^1$ instead of $3d^94s^2$. This is because the energy gap between the 3d and 4s orbitals is small enough to allow this electron promotion.

The outermost electronic configuration is $7s^1$. Elements ending in 's' orbitals are s-block elements. The number of electrons in the outermost 's' orbital determines the group for s-block elements. Since there is 1 electron in the $7s$ orbital, it belongs to Group 1. This configuration corresponds to Francium (Fr, Z=87).

Helium strictly belongs to the s-block as its configuration is $1s^2$. However, its positioning in the p-block (Group 18) along with other noble gases is justified because it has a completely filled valence shell ($1s^2$) and exhibits properties characteristic of other noble gases. Hydrogen, with $1s^1$, can be placed in Group 1 (alkali metals) or Group 17 (halogens) due to its ability to lose or gain an electron, respectively.

P-block elements are characterized by the filling of p-orbitals. Their general outer electronic configuration ranges from $ns^2np^1$ (Group 13) to $ns^2np^6$ (Group 18), excluding Helium which is an exception placed in Group 18.