Chemistry MCQs for NEET — Practice Questions with Answers

The NCERT states, 'The –OH group in alcohols and phenols is involved in intermolecular hydrogen bonding... It is interesting to note that boiling points of alcohols and phenols are higher in comparison to other classes of compounds, namely hydrocarbons, ethers and haloalkanes of comparable molecular masses.' Intermolecular hydrogen bonding requires more energy to overcome, leading to higher boiling points.

As per the NCERT text, 'Benzylic alcohols: In these alcohols, the —OH group is attached to a $sp^3$—hybridised carbon atom next to an aromatic ring.'

The NCERT defines vinylic alcohols as compounds where the –OH group is bonded to a carbon-carbon double bond, i.e., to a vinylic carbon or to an aryl carbon. The example given is $CH_2 = CH – OH$.

According to Table 7.2 in the NCERT, the common name for $CH_3OCH_3$ is Dimethyl ether.

The NCERT states, 'In fact, alcohols and phenols are Brönsted acids i.e., they can donate a proton to a stronger base (B:).'

The NCERT states, 'Ethers are classified as simple or symmetrical, if the alkyl or aryl groups attached to the oxygen atom are the same... Diethyl ether, $C_2H_5OC_2H_5$, is a symmetrical ether'.

The NCERT explains, 'The acidic character of alcohols is due to the polar nature of O–H bond. An electron-releasing group (–CH3, –C2H5) increases electron density on oxygen tending to decrease the polarity of O-H bond. This decreases the acid strength. For this reason, the acid strength of alcohols decreases in the following order: Primary > Secondary > Tertiary.'

The context states the general order of filling as 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s... After 3p, 4s is filled next. This is also consistent with the (n+l) rule where (n+l) for 3p is (3+1)=4 and for 4s is (4+0)=4. When (n+l) values are equal, the orbital with the lower 'n' value has lower energy. However, in the given sequence, 4s (n=4, l=0) has higher n but is filled before 3d (n=3, l=2). The general order provided in the NCERT text specifies 4s before 3d, making 4s the correct answer. (Reference: 'However, following order of energies of the orbitals is extremely useful: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s...')

The Pauli Exclusion Principle states that 'only two electrons may exist in the same orbital and these electrons must have opposite spin.' Therefore, the statement that electrons in the same orbital must have the same spin is incorrect. (Reference: 'According to this principle : no two electrons in an atom can have the same set of four quantum numbers. Pauli exclusion principle can also be stated as : “ only two electrons may exist in the same orbital and these electrons must have opposite spin.”')

The context states: 'pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.' It further clarifies, 'pairing of electrons will start in the p, d and f orbitals with the entry of 4th, 6th and 8th electron, respectively.' Since d-subshell has 5 degenerate orbitals, each orbital will receive one electron (total 5 electrons) before pairing starts. The 6th electron will be the first one to pair. (Reference: 'pairing of electrons will start in the p, d and f orbitals with the entry of 4th, 6th and 8th electron, respectively.')