P-block elements practice · Easy

Option D is correct because heating a mixture of $N{H_4}Cl$ and $NaN{O_2}$ produces pure nitrogen gas.

Option B ($s p^3 d$ and tetrahedral) is incorrect. The correct answer is $s p^3 d$ and trigonal bipyramidal because group 16 tetrafluorides have $s p^3 d$ hybridization and a see-saw or distorted tetrahedral structure, which is derived from a trigonal bipyramidal arrangement.

Option C is correct because liquid oxygen is pale blue in color.

Option C (All but fluorine show positive oxidation states) is the correct answer because fluorine is the most electronegative element and always exhibits a -1 oxidation state.

Option C (Neon) is correct because Neon (Ne) exists as a monoatomic gas, unlike the diatomic molecules of oxygen, fluorine, and nitrogen.

Option B (Square pyramidal) is correct because $XeOF_4$ has one lone pair and five bonding pairs, leading to a square pyramidal geometry.

Option C ($\mathrm{By\;FeS} + \mathrm{dil}.\mathrm{\;}{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$) is correct because Kipp's apparatus is designed to produce H2S gas by reacting ferrous sulfide (FeS) with dilute sulfuric acid.

Option B ($HIO_4 > HBrO_4 > HClO_4$) is not correct because the acidic strength of perhalic acids decreases down the group due to decreasing electronegativity of the central atom.

Option B ($\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}$) is correct because ammonium nitrate is an oxidizer that enhances the explosive power of TNT.

Option D (36 N) is correct because the normality of pure sulfuric acid is approximately 36 N, calculated using its density and molar mass.

Option C (The stability of hydrides increases from $N{H_3}\mathrm{ to }Bi{H_3}$ in Group 15) is correct because the stability of hydrides decreases down the group due to decreasing bond strength.

Option D (${\mathrm{N}_2}{\mathrm{O}_5}$) is correct because dinitrogen pentoxide ($N_2O_5$) is formed by removing water from two molecules of nitric acid ($HNO_3$).

Option C (${H_2}{S_2}{O_8}$) is correct because peroxodisulfuric acid (${H_2}{S_2}{O_8}$) contains a peroxide linkage (-O-O-).

Option C ($\mathrm{III} > I > II$) is correct because the acidity depends on the number of oxygen atoms bonded to the central atom; more oxygen atoms lead to higher acidity.

Option B (Reducing properties) is correct because ${\mathrm{H}_2}\mathrm{S}$ can donate electrons and get oxidized, thus acting as a reducing agent.

Option D is correct because ${N_2}{O_3}$ is a blue solid and is acidic in nature.

Option A ($\mathrm{N}{\mathrm{H}_3}$) is correct because in ammonia ($NH_3$), the oxidation state of nitrogen is -3, which is the lowest among the given options.

Option D is correct because in the reaction $2HN{O_3} + {P_2}{O_5} \to \mathrm{ }2HP{O_3} + {N_2}{O_5}$, $HN{O_3}$ acts as a dehydrating agent, not an oxidizing agent.

Option A ($NO$) is correct because nitric oxide ($NO$) has an odd number of electrons, resulting in an unpaired electron and making it paramagnetic.

Option D (HF is a stronger acid than HCl) is not true because HF is a weak acid due to strong hydrogen bonding, while HCl is a strong acid.

Option C (${N_3}H$) is correct because in hydrazoic acid ($N_3H$), the oxidation state of nitrogen is -1/3, which is higher than in the other options.

Option A is correct. Oxygen (O2) is a paramagnetic molecule due to the presence of unpaired electrons in its molecular orbital configuration.

Option B (${}_{88}^{226}\mathrm{Ra}$) is correct because Radium-226 decays through alpha decay, producing Radon (an inert gas) and then further decays to other inert gases.

Option A ($CO_2$) is correct because it reacts with CaO (a basic oxide) to form $CaCO_3$, but does not react with $NaHCO_3$ (a weak base).

Option B ($Ca{C_2} + C{a_3}{P_2}$) is correct because the mixture of calcium carbide and calcium phosphide produces acetylene and phosphine, which ignite spontaneously and produce a bright light.

Option D (${\mathrm{I}_2} < \mathrm{B}{\mathrm{r}_2} < C{\mathrm{l}_2} < {\mathrm{F}_2}$) is correct because oxidizing power increases as you move up the halogen group due to increasing electronegativity and decreasing bond dissociation energy.

Option A ($Ca_3P_2$) is correct because it reacts with water to produce phosphine ($PH_3$), which has a rotten fish smell and reacts with $CuSO_4$ to form $Cu_3P_2$.

Option A (${N_2}O$) is correct because ammonium nitrate decomposes upon heating to produce nitrous oxide (laughing gas) and water.

Option C is correct. Chlorine is used in the extraction of both Platinum (Pt) and Gold (Au).

Option B (Two moles of phosphine) is correct because calcium phosphide ($Ca_3P_2$) reacts with water to produce calcium hydroxide and phosphine ($PH_3$).