Classification of elements and Periodicity in properties MCQs for NEET — Chemistry Questions with Answers

The oxidation state in $N_{2}O <NO<N_2{O}_3<N_2{O}_4<N_2{O}_5$ dereases in the order +1<+2<+3<+4<+5.

In fact, N₂O and NO are neutral oxides.

The stability of the highest oxidation states decreases down the group and Bi does not form a pentoxide.Moreover higher oxidation states are more acidic than lower oxidation states.

Flourine reacts with dilute NaOH to form fluoride and oxygen fluoride

$$\begin{gathered} 2F_2 +2NaOH \to 2NaF +O{F}_2+H_{2}O \\ with hot and conc. NaOH, O_2 is liberated \\ 2F_2 +4NaOH\to 4NaF +O_2 +2H_{2}O \end{gathered}$$

tendency to get oxidise oxyacids of chlorine follows the order :

HClO >HClO₂ > HClO₃ > HClO₄ 

As we move from F^- to I^-, the size of the ion increases and hence electron density decreases from F^- to I^-. As a result,the tendency to attract proton decreases.Thus basic character decreases in the order F^- >Cl^- >Br^- > I^-.

Strongest acid possess smallest pK_a value.

N forms NCl₃, N₂O₅ and Ca₃N₂ . Nitrogen, due to no empty d- orbitals, cannot extend its covalency by more than three and hence do not form NCl₅.Due to its small size and high electronegativity it can accept 3 electrons to form $N^{3-} ion \left(C{a}_3{N}_2\right)$

Because of absence of d-orbitals, oxygen cannot increase its co-ordination number. Oxygen exhibits positive oxidation states of +1 in O₂F₂ and +2 in OF_2.

acidity of hydride increase down the group, due to increase in bond length

As per Table 3.4 and 3.5 in the NCERT text, the numerical roots for the digits 1, 0, and 4 are 'un', 'nil', and 'quad' respectively. Combining these in order of the atomic number digits (1-0-4) gives 'Unnilquad', and adding 'ium' at the end makes it Unnilquadium.