NEET Practice Questions (MCQs) with Answers & Solutions

The NCERT text explicitly mentions, 'In fact the behaviour of lithium and beryllium is more similar with the second element of the following group i.e., magnesium and aluminium, respectively. This sort of similarity is commonly referred to as diagonal relationship in the periodic properties.'

The NCERT text explains, 'In addition, the first member of group has only four valence orbitals (2s and 2p) available for bonding, whereas the second member of the groups have nine valence...'. This is due to the absence of d-orbitals in the second period elements (n=2, so only 2s and 2p orbitals are available).

The context states, 'In fact the behaviour of lithium and beryllium is more similar with the second element of the following group i.e., magnesium and aluminium, respectively. This sort of similarity is commonly referred to as diagonal relationship in the periodic properties.' Therefore, Li and Mg exhibit a diagonal relationship.

The first element of group 13 (Boron) exhibits anomalous properties due to its small size, large charge/radius ratio, and high electronegativity (II). Like other second-period elements, it only has 2s and 2p orbitals for bonding (III), leading to limited valency and orbital expansion. Unlike its heavier congeners which tend to form more ionic compounds, Boron forms compounds with more covalent character due to its high polarizing power. Thus, statement I is incorrect.

The NCERT states, 'For example, lithium unlike other alkali metals... form compounds with pronounced covalent character; the other members of these groups predominantly form ionic compounds.' Therefore, Liâ‚‚O would exhibit more covalent character compared to Naâ‚‚O, which is predominantly ionic.

The NCERT states that 'lithium unlike other alkali metals, and beryllium unlike other alkaline earth metals, form compounds with pronounced covalent character'. This anomalous behavior, including covalent character, is attributed to their 'small size, large charge/radius ratio and high electronegativity'.

The text states, 'the first member of group has only four valence orbitals (2s and 2p) available for bonding'. This means that second-period elements cannot expand their octet, thus restricting their maximum covalency to four (one s and three p orbitals). The absence of d-orbitals prevents higher covalency.

According to the NCERT, 'The anomalous behaviour is attributed to their small size, large charge/radius ratio and high electronegativity of the elements.' These factors lead to a greater tendency for covalent bonding in second-period elements compared to their heavier group members (Y), which tend to form more ionic compounds due to larger size and lower electronegativity. Atomic radius increases down a group, and electronegativity decreases down a group. Second-period elements have fewer valence orbitals (2s, 2p) than third-period elements (3s, 3p, 3d, where 3d orbitals can be utilized).

The NCERT guidelines for solving problems in mechanics state: 'Do not include the forces on the environment by the system' when drawing a free-body diagram. This means the free-body diagram should only show forces acting ON the chosen system.

When two forces act perpendicularly, their resultant force (F_net) can be found using the Pythagorean theorem: $F_{net} = \sqrt{(F_1)^2 + (F_2)^2} = \sqrt{(8N)^2 + (6N)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 N$. According to Newton's Second Law ($F_{net} = ma$), the acceleration $a = F_{net}/m = 10 N / 5 kg = 2 m/s^2$.