Chemistry MCQs for NEET — Practice Questions with Answers

Table 7.2 lists the compound $\text{C}_6\\text{H}_5\\text{O(CH}_2\\text{)}_6\\text{CH}_3$ as 'Heptyl phenyl ether' by common name and '1-Phenoxyheptane' by IUPAC name. The larger group is chosen as the parent hydrocarbon (heptane), and the smaller group with oxygen becomes the alkoxy substituent (phenoxy).

The text explicitly states: 'The simplest hydroxy derivative of benzene is phenol. It is its common name and also an accepted IUPAC name.'

Benzylic alcohols are those in which the —OH group is attached to an $\text{sp}^3$-hybridised carbon atom next to an aromatic ring. $\text{C}_6\\text{H}_5\\text{CH}_2\\text{OH}$ (benzyl alcohol) fits this description. $\text{CH}_2\\text{=CH-OH}$ is a vinylic alcohol, $\text{CH}_3\\text{CH(OH)CH}_3$ is a secondary propanol, and $\text{CH}_2\\text{=CH-CH}_2\\text{OH}$ is an allylic alcohol.

According to the NCERT text, 'p orbitals increase in size and energy with increase in the principal quantum number and hence the order of the energy and size of various p orbitals is 4p > 3p > 2p.' Therefore, statement o3 is incorrect.

The NCERT text states, 'The shapes of the first four d-orbitals are similar to each other, where as that of the fifth one, $d_{z^2}$, is different from others, but all five 3d orbitals are equivalent in energy.'

The NCERT text states, 'The number of nodes are given by the n –2, that is number of radial node is 1 for 3p orbital, two for 4p orbital and so on.' This formula refers to radial nodes specifically for p-orbitals (where l=1). The general formula for radial nodes is $n-l-1$, so for p orbitals, it is $n-1-1 = n-2$.

The NCERT text explicitly states, 'According to the German physicist, Max Born, the square of the wave function (i.e., $ψ^2$) at a point gives the probability density of the electron at that point.'

The NCERT text says, 'It may be noted that for 1s orbital the probability density is maximum at the nucleus and it decreases sharply as we move away from it.'

The NCERT text states, 'It is also observed that the size of the s orbital increases with increase in n, that is, 4s > 3s > 2s > 1s and the electron is located further away from the nucleus as the principal quantum number increases.'

The NCERT text explains, 'Boundary surface diagram for a s orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. It encloses a region in which probability of finding the electron is about 90%. Thus, we see that 1s and 2s orbitals are spherical in shape. In reality all the s-orbitals are spherically symmetric.'