Option (D) is correct.
The valency of carbon is 4 but one carbon has three bonds and other five, whereas each should have four bonds.

From its structure it can be seen that, number of electron pairs in $$XeF_4$$ at the corners of the square are four

$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 },3{ s }^{ 2 }3{ p }^{ 1 }$$
As it is p block element therefore the group is $$12+1=13$$ (III A group)

For p-dichlorobenzene and p-dicyanobenzene, the dipole moments of individual bonds cancel each other as they are equal in magnitude and opposite in direction. Hence, the resultant dipole moment of the molecules is zero. For p-hydroquinone and p-benzenedithiol, the $$O-H$$ and $$S-H $$ bond dipoles in these molecules do not cancel each other as they are not in opposite directions due to existence in different conformations. Hence, the resultant dipole moment of the molecules is non zero.

The maximum covalency of B is 4. It cannot form 6 bonds because it has no vacant orbital in the outermost orbit.

$$C_2^{2-}$$ ion $$\longrightarrow$$ $$[C\equiv C]^{2-}$$

The structure of HCN is $$H-C\equiv N$$. It has $$2 \pi$$ and $$2 \sigma $$bonds.

$$CCl_4  < CHCl_3 < CH_2Cl_2 <CH_3Cl$$

Hint: Those species which have the same number of electrons and identical shapes are called isoelectronic and isostructural species.

Correct Answer: Option C and D

Explanation:

Compound	Number of electrons	Hybridization	Shape	Structure
$$SO_3^{2-}$$	$$16+2+24=42$$	$$sp^3$$	Tetrahedral
$$NO_3^-$$	$$7+24+1=32$$	$$sp^2$$	Trigonal planar
$$ClO_3^{-}$$	$$17+1+24=42$$	$$sp^3$$	Tetrahedral
$$CO_3^{2-}$$	$$6+24+2=32$$	$$sp^2$$	Trigonal planner

 

Final Answer: 

$$NO_3^-$$ and $$CO_3^{2-}$$ are isoelectronic and isostructural. 

$$SO_3^{2-}$$ and $$ClO_3^{-}$$ are isoelectronic and isostructural.

$$(n+1)$$ values for ,$$4d=4+2=6$$
$$5p=5+1=6$$
$$5f=5+3=8$$
$$6p=6+1=7$$
$$\therefore$$ Correct order of energy would be $$5f>6p>5p>4d$$

$$\displaystyle CN^{-}, CO$$ are isoelectronic species (contains same number of electrons). They have same bond order.

Chlorine reacts more rapidly with $$H_2$$ to form HCl which having an ionic bond. An ionic bond is stronger than a covalent bond.

Solution:- (B) Antiferromagnetic

When two atoms of the elements having the same spin of electron approach for bonding, the orbital overlap will not occur and bonding will not occur. Hence, Both $$(A)$$ and $$(B)$$ are correct.
However, if two atoms of the elements having opposite spin of electron approach for bonding, the orbital overlap will occur and bonding will not occur.

$$O=C=O$$ , each oxygen shares 2 electron with carbon which makes the covalency 4 

The molecules $$A, C & D$$ are alkene, alkyne and alkadiene. They contain both $$\sigma$$ and $$\pi$$ bonds. Alkanes undergo $$sp^3$$ hybrid orbitals and do not contain any $$\pi$$ bonds. They contain only $$\sigma$$ bonds.

For P block elements, maximum covalency = maximum number of electrons in the valence shell = group number. Sulphur is in group XVI. So, it will have a maximum covalency of 6. 

Concept of valency is the simple way of counting electrons and forming bonds and it does not mean that the given atom has to form bonds with the neighbors and in organic chemistry carbon forms four single bonds to neighbors so it has 4 valency.

Covalency is the process of attaining inert gas configuration by sharing of electrons.

If an element has some valency that is, for example, $$N - 1s^{2}2s^{2}2p^{3}$$ here it has the combining the capacity of 3 that is it can share 3 electrons or accept three electrons. Here even valency is nothing but 3. Valency is defined as combining the power of the element with other elements.

$$B^{3-} - 1s^22s^22p^4$$
incomplete octet

$$ \mathrm SF_{4}: sp^{3}d $$
$$1$$ Lone pair of electrons.

Valence shell electron pair repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules from the number of electron pairs surrounding their central atoms.The premise of VSEPR is that the valence electrons pairs surrounding an atom tend to repel each other, and will therefore adopt an arrangement that minimizes this repulsion, thus determining the molecule's geometry.

The two main ways in which atoms can be combined to form molecules are by electrovalent bonds and covalent bonds.Electrovalent bonds are held together by an electrical attraction between positively charged cations and negatively charged anions.

The Valence Bond theory describes the formation of covalent bonds from the overlap of atomic orbitals on two different atoms. Because of the overlap, it is highly probable that a pair of electrons are found in the physical region or space where the orbitals overlap.

The covalency of an atom is the number of electrons it shares with other atoms when forming chemical bonds. So covalency of oxygen is two as it forms a double bond and that of nitrogen is three as it forms a triple bond.

Out of the given options, van der waal forces are the weakest.

A  triple bond can be considered as one sigma bond and two pi bonds. The head-on overlap between two $$p_{x}$$ orbitals of two-atom forms the sigma bond. The lateral/ or sideway overlap between $$p_{y}$$ and $$p_{z}$$ orbitals forms two $$\pi $$ bonds. So the correct option is B

Covalent bond is formed by the sharing of electrons between the atoms and the phenomenon is known as covalency.

Solution:- (C) The link between atom

Chemical bond formation is always an exothermic process and chemical bond breaking is an endothermic process.

Ionic bond is the bond in which electrons are transferred.

Hint: Covalency is equal to the number of bonds

$$Be$$ has $$2$$ electrons in the valence shell (it belongs to group $$2$$ of the periodic table). 

Ethane consists of $$6\:C-H$$ bonds and $$ 1\: C-C$$ bond. Total number of bonds is  $$7$$. Each bond is made up of two electrons. Thus, total bonded electrons are $$7\times2$$ = $$14$$.

Xenon difluoride is $$XeF_{2}$$. $$Xe$$ is a noble gas. So, it has $$8 $$ electrons in the valence shell. Out of which, $$2$$ are involved in formation of covalent bond with the $$2 \:F$$ atoms. So, $$6$$ electrons are left as unbonded. Therefore, lone pairs = half of $$6$$ = $$3$$.

$$Xe$$ is a noble gas. It has $$8$$ electrons in its valence shell. Now, it is bonded to $$4\: F$$ atoms. 

Chemical bond is an attraction between atoms that allows the formation of chemical substances. The bond will be formed where attraction forces due to opposite charges were balanced by repulsion forces due to similar charges.

 $$Cl^{\cdot} + Cl^{\cdot}  \longrightarrow  Cl : Cl   $$

Urea has maximum covalent bonds - $$8$$.

Covalency of an atom in a covalent compound is the number of covalent bonds formed by an atom. Maximum covalency of Nitrogen is 4 (example - ammonium ions).

Lone pairs are in the order: 3, 2, 1 and 0 respectively as shown in the figure

Bent or angular shape. When drawing a Lewis Dot structure, the central atom has two elements (two hydrogen) attached to it and the central atom (oxygen) has two lone pairs which lead to the formation of a bent shape.

(a)  Ethane has $$6 \:C-H$$ bonds and $$1 \:C-C$$ bond.
(b) Ammonia has $$3\: N-H$$ bonds.
(c) Sulphur hexafluoride has $$6\: S-F$$ bonds.
(d) Bromine pentafluoride has $$5\: Br-F$$ bonds.