p-Block Elements

In graphite, carbon is $s{p^2}$ hybridized. Each carbon is thus linked to three other carbon atoms forming hexagonal rings. Since only three electrons of each carbon are used in making hexagonal ring, fourth electron of each carbon is free to move. This makes graphite a good conductors of heat and electricity. Further graphite has a two dimensional sheet like structure. These various sheets are held together by weak van der Waal's force of attraction. due to these weak forces of attraction, one layer can slip over the other. This makes graphite soft and a good lubricating agent.

When phosphine gas (PH₃) is mixed with chlorine gas (Cl₂), the reaction between them depends on the conditions and the proportions of the reactants. Let's consider the options :

1. Option A : PCl₃ and HCl are formed and the mixture warms up.

   
   
   • When phosphine gas reacts with chlorine gas, one of the possible products is phosphorus trichloride (PCl₃) along with hydrogen chloride (HCl). This reaction is exothermic, meaning it releases heat and causes the mixture to warm up.



2. Option B : PCl₅ and HCl are formed and the mixture cools down.

   
   
   • Phosphorus pentachloride (PCl₅) can be formed if an excess of chlorine is used, but this reaction is also exothermic. The formation of PCl₅ is less likely with PH₃ as it requires a higher proportion of chlorine and controlled conditions.



3. Option C : PH₃·Cl₂ is formed with warming up.

   
   
   • There is no known stable compound like PH₃·Cl₂ formed directly from the reaction of phosphine and chlorine gas. This option does not represent a typical chemical reaction product.



4. Option D : The mixture only cools down.

   
   
   • Chemical reactions between phosphine and chlorine are exothermic, so this statement is incorrect.

Based on this analysis, the most likely and correct reaction is described in Option A : PCl₃ and HCl are formed, and the mixture warms up. This reaction is feasible and aligns with the known chemistry of phosphine and chlorine gases.

Among the given compounds, the $N{H_3}$ is most basic. Hence has highest proton affinity.

In the formation of $Xe{F_2},\,s{p^3}d\,\,$ hybridisation occurs which gives the molecule a trigonal bipyramidal structure.



In the formation of $Xe{F_4},\,s{p^3}{d^2}$ hybridization occurs which gives the molecule an octahedral structure.



In the formation of $Xe{F_6},\,s{p^3}{d^3}$ hybridization occurs which gives the molecule a pentagonal bipyramidal structure.

${}_7N = 1{s^2}2{s^2}\,2{p^3};\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{}_{15}P = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$

NOTE : In phosphorous the $3d$-orbitals are available. Hence phosphorous can from pentahalides also but nitrogen can not form pentahalide due to absence of $d$-orbitals.

When H₂S is passed through Hg₂S we get a mixture of mercurous sulphide and mercury(HgS + Hg₂S).

$CaOC{l_2} - $or it can also be written as $Ca\mathop {\left( {OCl} \right)}\limits_{{x_1}} \mathop {Cl}\limits_{{x_2}} $

hence oxidation no of $Cl$ in $OC{l^ - }$ is

$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2 + {x_2} = - 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} = 2 - 1 = + 1 \cr} $

now oxidation no. of another $Cl$ is $-1$ as it is present as $C{l^ - }.$

The correct answer is Option A: N: tetrahedral, sp³; B: tetrahedral, sp³

Explanation:

• BF₃ (Boron trifluoride): In its isolated state, BF₃ has a trigonal planar geometry around the boron atom with sp² hybridization.


• NH₃ (Ammonia): In its isolated state, NH₃ has a trigonal pyramidal geometry around the nitrogen atom with sp³ hybridization.

However, when BF₃ and NH₃ form a 1:1 complex, the lone pair of electrons on nitrogen donates to the empty p-orbital of boron. This leads to the formation of a coordinate bond between nitrogen and boron.

In the BF₃-NH₃ complex:

• Nitrogen (N): Now has four bonding pairs of electrons (three with hydrogen and one with boron), resulting in a tetrahedral geometry and sp³ hybridization.


• Boron (B): Also has four bonding pairs of electrons (three with fluorine and one with nitrogen), leading to a tetrahedral geometry and sp³ hybridization.

Key point: The formation of the coordinate bond between nitrogen and boron changes the hybridization and geometry of both atoms in the complex.

The oxidation number of an atom in a molecule or an ion is essentially a theoretical charge that the atom would have if all the bonds to atoms of different elements were completely ionic. To determine the oxidation numbers of sulfur in the given compounds: S₈, S₂F₂, and H₂S, let's analyze them one by one.

1. In S₈: This is an elemental form of sulfur, comprising of sulfur atoms only. In elemental forms, the atoms are not bonded to atoms of a different element, but only to themselves. In such cases, the oxidation number is always 0. Hence, the oxidation number of sulfur in S₈ is 0.

2. In S₂F₂: This is a molecule consisting of sulfur and fluorine. Fluorine is more electronegative than sulfur and virtually always has an oxidation number of -1 in its compounds. Let's denote the oxidation number of sulfur as $x$. Since there are two fluorine atoms, each contributing -1, and the overall molecule is neutral, we get:

$2x + (2 \times -1) = 0$

$2x - 2 = 0$

$2x = 2$

$x = +1$

Therefore, the oxidation number of sulfur in S₂F₂ is +1.

3. In H₂S: In this compound, hydrogen has an oxidation number of +1 (except when bonded to metals where it can be -1). Since there are two hydrogens, their total contribution is +2. Let's denote the oxidation number of sulfur in this compound as $x$. The compound is neutral overall, so:

$2(+1) + x = 0$

$2 + x = 0$

$x = -2$

Therefore, the oxidation number of sulfur in H₂S is -2.

Comparing with the given options, the correct answers are 0, +1, and -2 for S₈, S₂F₂, and H₂S, respectively. Therefore, the correct option is Option A: 0, +1, and -2.