The p-Block Elements Test - 10

From the electronic configuration given, it is clear that element \(\mathrm{M}\) is 1 short of octet configuration. It belongs to group 17. Group 17 elements exist as diatomic molecules.

Hence, the correct option is (B).

Ammonia acts as a Lewis base due to the presence of lone pair on the nitrogen atom. So, it can form coordinate bond transition metal cations to form complex compounds. Ammonia reacts with a solution of \(C u^{2+}\) to from a deep blue coloured complex. 

The reaction for the above complex compound can be written as follows:

\(C u^{2+}(a q)\)[Deep Blue] \( + 4 N H_{3}(a q) \longrightarrow\left[C u\left(N H_{3}\right)_{4}\right]^{2+}(a q)\)[Deep Blue]

Hence, the correct option is (A).

In cyclo- \(S_{6}\) molecule, the ring adopts chair form as shown in the figure.

Rhombic sulphur is yellow in colour.

Rhombic sulphur cannot be dissolved in water but can be dissolved in benzene, ether, alcohol etc.

Both rhombic and monoclinic sulphur have the molecular formula \(S_{8}\). 

Hence, the correct option is (D).

In the mixture of nitrite and nitrate ions, the nitrite ions can be removed by heating with urea and supluric acid.

The reaction takes place as follows:

\(2 \mathrm{NaNO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{HNO}_{2}\)

\(2 \mathrm{HNO}_{2}+\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{N}_{2}+3 \mathrm{H}_{2} \mathrm{O}\)

In the above reaction, we can see that the products, nitrogen and carbon dioxide are in gaseous form and thus, do not stay in the solution. This reaction is exothermic and thus, will occur spontaneously without the supply of any extra energy.

Hence, the correct option is (C).

Atomicity of an element is a measure of total number of atoms present in a molecule. 

It is determined by the ratio of molecular mass of total number of atoms to atomic mass of an element.

For phosphorus,

Atomic mass \(=30.97\approx 31\)

Since, there are 4 phosphorous atoms per molecule.

Therefore,

Molecular mass \(=4\times31=124\)

Therefore, 

Atomicity \(=\frac{124}{31}=4\)

Hence, the correct option is (D).

A clathrate is a crystalline substance consisting of a lattice that traps or contains a molecule.

Clathrate of Xenon with water is the lattice consists of \(\mathrm{X e}\) as the trapped molecule.

There is no coordinate bond present between \(\mathrm{X e}\) and \(\mathrm{H_{2} O}\); rather dipole-induced dipole interaction present.

Hence, the correct option is (D).

The structure of \(P_{4} S_{3}\) is as follows:

Then, the total number of \(P-S\) bonds are 6.

Hence, the correct option is (C).

\(\mathrm{Br}\)-atom has configuration:

\(1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{10}, 4 s^{2} 4 p^{5}\)

• To get pentavalency, two of the p-orbitals are unpaired and electrons are shifted to \(\mathrm{4 d}\)-orbitals. 
• In this excited state, \(sp^3 d^2\)-hybridization occurs giving octahedral structure. 
• Five positions are occupied by \(\mathrm{F}\) atoms forming sigma bonds with hybrid bonds and one position occupied by lone pair, i.e., the molecule as a square pyramidal shape.

So, from the square pyramidal structure of \(\mathrm{B r F_{5}}\) as shown above, four fluorine atoms are present in the same plane. However, the Bromine, central atom lies slightly below the square planer.

Hence, the correct option is (C).

In solid state \(N_{2} O_{5}\) exists as \(\left[\mathrm{NO}_{3}\right]^{-}\left[\mathrm{NO}_{2}\right]^{+}\).

• Solid \(N_{2} O_{5}\) is a salt, consisting of separated anions and cations. 
• The cation is the linear nitronium ion \(\mathrm{NO}_{2}^{+}\)and the anion is the planar nitrate \(\mathrm{NO}_{ {3}}^-\) ion. 
• Thus, the solid could be called nitronium nitrate. 
• Both nitrogen centers have oxidation state \(+5\).

Hence, the correct option is (A).

Since nitrogen atom have 5 electrons in its outermost shell, so higher electronegative elements can extend its oxidation state up to \(+5\), while in case of taking electrons it cannot go beyond 8 electrons. So, at most it can accept 3 electrons.

Hence, the correct option is (A).

When dissociation increases, the percentage of nitrogen dioxide, will also increase. While vapour density is decreased, it means vapour density is inversely proportional to the dissociation.

Therefore, 

The dissociation vapour density will be high in following order:

\(d_{25\%}>d_{50\%}>d_{75\%}>d_{100\%}\)

Therefore, the required sequence will be:

\(d_{1}>d_{2}>d_{3}>d_{4}\)

Hence, the correct option is (A).

The acid formed by diphosphorus trioxide i.e., \(\mathrm{P_{2} O_{3}}\) is \(\mathrm{H}_{3} \mathrm{PO}_{3}\). When diphosphorus trioxide i.e., \(\mathrm{P}_{2} \mathrm{O}_{3}\) reacts with water then \(\mathrm{H}_{3} \mathrm{PO}_{3}\) i.e., phosphorus acid will form. 

The reaction is as follows: 

\(2 \mathrm{P}_{2} \mathrm{O}_{3}+6 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{3}\).

Hence, the correct option is (D).

The total number of electrons that take part in forming bonds in \(\mathrm{N}_2\) is \( 6\).

Structure of \(N_{2}\) is \(N \equiv N\).

• We know that the atomic number of nitrogen is 7 and the electronic configuration of nitrogen is 2, 5 which means two electrons are present in s-orbital and five electrons in p-orbital.
• When two nitrogen atoms will react, three electrons (from five valence electrons) of each nitrogen atom are shared between them so that both will obtain a stable configuration.
• In \(N_{2}\) there are 3 bonds between two nitrogen atoms and each bond is formed by 2 electrons.

The full outer shells with the shared electrons are now stable forming a covalent bond and this can be shown through the following diagram:

So, 3 bonds make it \(3 \times 2=6\) electrons in total.

Hence, the correct option is (C).

Elements of the group-15 form compounds in +5 oxidation state. However, bismuth forms only one well-characterized compound in +5 oxidation state. The compound is \(\mathrm{B i F_{5}}\).

• Due to inert pair effect bismuth only exhibit \(+3\) oxidation state and forms trihalides only. 
• But due to small size and high electronegativity of Florine Bismuth forms \(\mathrm{B i F_{5}}\) only.

Hence, the correct option is (B).

VIA group or Group 16 elements belong to the oxygen family are also called chalcogens. 

• The name is derived from the Greek word 'chalcos', which means 'ore' and 'gen', which means 'formation'.
• They are called so because most of the copper ores have copper in the form of oxides and sulphides. 
• They also contain small amounts of selenium and tellurium.

Hence, the correct option is (A).

The reaction of decomposition of phosphine can be written as:

\(\mathrm{4 P H_{3} \rightarrow P_{4}+6 H_{2}}\)

The balanced chemical reaction indicates that four moles of phosphine generates one mole of phosphorus and six moles of hydrogen.

The phosphorus exists as a triatomic molecule in elemental form. It is a solid and therefore, its volume change is negligible according to gas laws.

The volume of phosphine given is \(100 \mathrm{~mL}\). The amount of hydrogen produced can be calculated as:

Amount of hydrogen \(=\frac{6}{4} \times\) amount of phosphine

\(=\frac{6}{4} \times 100 \mathrm{~mL}\)

\(=150 \mathrm{~mL}\)

Thus, \(100 \mathrm{~mL}\) of phosphine gas generates \(150 \mathrm{~mL}\) of hydrogen gas.

The change in volume \(=150-100\)

\(=50 \mathrm{~mL}\)

Hence, the correct option is (A).

Generally, crops absorb nitrogen in the form of nitrate because crops cannot digest the atmospheric nitrogen due to the presence of the triple bond in the atmospheric nitrogen.

Most crops preferably absorb nitrogen in nitrate form. 

• Nitrate fertilizers (like potassium nitrate) are readily soluble in water and meet the immediate nitrogen requirement of plants. 
• Nitrogen is involved in forming amino-acids, which are the building blocks of proteins.
• Nitrate form of fertilizer is more soluble in water than the other forms of the fertilizer.
• Among the given options potassium nitrate is the fertilizer which is highly soluble in water.

Therefore, potassium nitrate is a nitrogen fertilizer that can meet the needs of the immediate requirement to the crops.

Hence, the correct option is (A).

In the reaction of potassium dichromate with hydrogen sulphide in presence of an acidic medium of sulphuric acid, a redox reaction takes place as follows:

During the reaction, the potassium chromate acts as the oxidizing agent, due to the presence of the hexavalent chromium in (+6) oxidation state. It is thus reduced to (+3) oxidation state.

Then, the reduction half-cell reaction will be:

\(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}+14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)

Also, the sulphide \(S^{2-}\) acting as the reducing agent, is oxidised to sulphur (S) in the reaction. Then, the oxidation half-cell reaction will be:

\(\mathrm{S^{2-} \longrightarrow S+2 e^{-}}\)

Thus, combining the oxidation and reduction half-cell reaction, the balanced equation obtained is:

\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+3 \mathrm{H}_{2} \mathrm{~S}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3 \mathrm{S}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+7 \mathrm{H}_{2} \mathrm{O} \)

The products formed are sulphur, potassium sulphate, chromium sulphate and water.

Hence, the correct option is (A).

\(\mathrm{P C l_{3}}\) or phosphorous trichloride is a chemical compound which readily reacts with water to produce \(\mathrm{H C l}\) gas, and with \(\mathrm{H C l}\) gas the by-product formed is \(\mathrm{H_{3} P O_{3}}\). 

The reaction is as follows:

\(\mathrm{PCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}+3 \mathrm{HCl}\) 

Therefore, the products formed by the hydrolysis of \(\mathrm{P C l_{3}}\) are \(\mathrm{H}_{3} \mathrm{PO}_{3}\) and \(\mathrm{HCl}\).

Hence, the correct option is (A).

Ozone \((\mathrm{O_{3}})\) is also one of the greenhouse gas.

• The contribution of \(\mathrm{O_{3}}\) to the greenhouse effect is about 8 to \(10 \%\). 
• About \(75 \%\) of solar energy is absorbed by the surface of the earth, and the rest is radiated back to the atmosphere. 
• This heat traps gases like \(\mathrm{CO}_{2}, \mathrm{CH}_{4}, \mathrm{CFCs}\) and \(\mathrm{H}_{2} \mathrm{O}\) present in the atmosphere and adds to the heat of the atmosphere, causing global warming.

Hence, the correct option is (A).

Nitrogen is an inactive element as it has a high bond dissociation energy due to the presence of triple bond \(N \equiv N\). Therefore, viewed as a non-reactive element.

Hence, the correct option is (D).

The difference in the bond angles is linked with the electronegativities of oxygen and sulphur atoms. As we move down the group from \(\mathrm{O}\) to Te, the size of the central atom goes on increasing and its electronegativity goes on decreasing.

As a result, the position of the two bond pairs shifts away and away from the central atom as we move from \(\mathrm{H}_{2} \mathrm{O}\) to \(\mathrm{H}_{2} \mathrm{Te}\). Consequently, the repulsion between the bond pairs decreases as we move from \(\mathrm{H}_{2} \mathrm{O}\) to \(\mathrm{H}_{2} \mathrm{Te}\) and therefore, the bond angle decreases in the order.

Hence, the correct option is (B).

The ions \(\mathrm{Mg}^{+2}\) and \(\mathrm{PO}_{4}^{-3}\) combine to form compound \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\). 

Total charge on cation \(=3(2+)=+6\). 

Total charge on anion \(=2(-3)=-6\).

Therefore, 

Total charge on cation \(=\) Total charge on anion.

Hence, the correct option is (A).

According to Le Chatelier's principle, high pressure would favour the formation of \(\mathrm{N H_{3}}\).

\(\mathrm{N_{2}}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \leftrightharpoons \mathrm{N H_{3}}(\mathrm{~g})\)

The optimum conditions for the production of \(\mathrm{N H_{3}}\) are pressure of \(200 \times 10\) \({ }^{5} \mathrm{~Pa}\) (about \(\left.200 \mathrm{~atm}\right)\) and temperature of \(-700 \mathrm{~K}\).

In the given reaction, the number of moles of reactant is more than the product and according to the ideal gas equation, the pressure is inversely proportional to the number of moles.

Thus, by increasing pressure i.e., at high pressure, the reaction will proceed in a forward direction towards less number of moles. Thus, the formation of the ammonia increases.

Hence, the correct option is (A).

(A) \(\mathrm{N H_{4} N O_{2}}\) is Ammonium Nitrite. The thermal decomposition occurs as follows-

\(\mathrm{NH}_{4} \mathrm{NO}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

The reaction produces Nitrogen gas and water vapour.

(B) \(\mathrm{N a N_{3}}\) is sodium azide. The thermal decomposition occurs as follows-

\(2 \mathrm{N a N_{3}} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{N a}+3 \mathrm{N_{2}}\)

The reaction produces Sodium metal and Nitrogen gas. Very pure nitrogen is obtained by this method.

(C) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is Ammonium dichromate. These are red coloured crystals. The thermal decomposition occurs as follows-

\(\left(\mathrm{NH}_{4}\right){ }_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{N}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

Ammonium dichromate is thermodynamically unstable and hence it decomposes into dark green chromium (III) oxide and Nitrogen gas. The reaction occurs with flashes of light.

We can see that, \(\mathrm{N}_{2}\) is released in all the reactions.

Hence, the correct option is (D).

We know that, electronegativity of halogen group decreases on moving down the group because size of atom increases. Decreasing order of electronegativity of halogens is F, Cl, Br, I.

In case of \(\mathrm{s p^{3}}\) hybridization:

• The axial position of metal has less s-character and the equatorial position of metal has more s-character.
• The ligand which are s-phile (s-lover) will tend to occupy equatorial position, and the ligands which are s-nonphile will tend to occupy axial position. 
• The s-phile ligand have less electronegativity or we can say s-nonphile (having more electronegativity) atoms have tendency to occupy axial position is called apicophilicity.
• Thus, \(\mathrm{C l^{-}}\)is more electronegative than \(\mathrm{B r^{-}}\).

Therefore, it is s-nonphile and will occupy axial position. Thus, the correct structure of \(\mathrm{P C l_{3} B r_{2}}\) will be as shown:

Hence, the correct option is (A).

The structure of white phosphorus is tetrahedral.

Phosphorus or s tetraphosphorus \(\left(P_{4}\right)\) exists as molecules made up of four atoms in a tetrahedral structure. The molecule is described as consisting of six single P-P bonds.

Hence, the correct option is (C).

As we move from top to bottom the acidic nature of hydrides of the group increases as the electronegativity of atom decreases.

So, the order of acidity of hydrides of the Vth group is:

\(\mathrm{H}_{2} \mathrm{O}<\mathrm{H}_{2} \mathrm{~S}<\mathrm{H}_{2} \mathrm{Se}<\mathrm{H}_{2} \mathrm{Te}\)

Strongest acid among the given is \(\mathrm{H}_{2} \mathrm{Te}\).

Hence, the correct option is (D).

Ammonia gas is produced when calcium nitride \(\left(\mathrm{Ca}_{3} \mathrm{N}_{2}\right)\) is hydrolyzed by water.

The reaction is given as follows:

\(\mathrm{Ca}_{3} \mathrm{N}_{2}+6 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{NH}_{3}\)

Thus, the gas formed is ammonia.

Hence, the correct option is (B).

Xenon (Xe) is inert though it gives compounds with Fluorine due to its large size and lower ionization potential as compared to other noble gases. Also, it is more reactive than other noble gases.

Hence, the correct option is (A).