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2 years ago

A molecule in which the central atom forms three single bonds and has one lone pair is said to have a ________ shape.

Chemistry

1 answer:

nignag [31]2 years ago

5 0

Answer : The correct option is, (c) pyramidal

Explanation :

As we are given that a molecule in which the central atom forms three single bonds and has one lone pair that means the central metal atom has 3 bond pairs and 1 lone pair of electrons.

There are total 4 electron pairs. So, the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are three atoms or bonds around the central atom and the fourth position occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be pyramidal.

Hence, correct option is, (c) pyramidal

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What happens when an atom of sulfur combines with two atoms of chlorine to produce SCl2?

RoseWind [281]

Answer:

Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom

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3 years ago

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

Westkost [7]

Answer:

a) $K_{2} S$ and $NH_{4} Cl$ :

There are no insoluble precipitate forms.

b) $Ca Cl_{2}$ and $(NH_{4} )_{2} Co_{3}$ :

There are the insoluble precipitates of $CaCo_{3}$ forms.

c) $Li_{2}S$ and $MnBr_{2}$ :

There are the insoluble precipitates of $MnS$ forms.

d) $Ba(No_{3} )_{2}$ and $Ag_{2} So_{4}$ :

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e) $Rb_{2}Co_{3}$ and $NaCl$ :

There are no insoluble precipitates forms.

Explanation:

Solubility rule suggests:- $K_{2} S$ ⇒ soluble, $NH_{4} Cl$ ⇒ soluble.

KCl ⇒ soluble, $(NH_{4})_{2} S$ ⇒ soluble.

There are no insoluble precipitate forms.

Solubility rule suggests:- $Ca Cl_{2}$ ⇒ soluble, $(NH_{4} )_{2} Co_{3}$ ⇒ soluble.

$CaCo_{3}$ ⇒ insoluble, $NH_{4} Cl$ ⇒ soluble.

There are the insoluble precipitates of $CaCo_{3}$ forms.

Solubility rule suggests:- $Li_{2}S$ ⇒ soluble, $MnBr_{2}$ ⇒ soluble.

$LiBr$ ⇒ soluble, $MnS$ ⇒ insoluble.

There are the insoluble precipitates of $MnS$ forms.

Solubility rule suggests:- $Ba(No_{3} )_{2}$ ⇒ soluble, $Ag_{2} So_{4}$ ⇒insoluble.

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

Solubility rule suggests:- $Rb_{2}Co_{3}$ ⇒ soluble, $NaCl$ ⇒ soluble.

$RbCl$ ⇒ soluble, $Na_{2} Co_{3}$ ⇒ soluble.

There are no insoluble precipitates forms.

6 0

2 years ago

A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the subs

Marrrta [24]

Question is incomplete, the complete question is as follows:

A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?

A. Toxicity, because it can be observed by altering the state of the substance

B. Boiling point, because it can be observed by altering the state of the substance

C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms

D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms

Answer:

Explanation:

A student can examine a substance without altering the bonds within the molecules by examining its boiling point.

The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.

Hence, the correct answer is "B".

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3 years ago

What do you understand by Gas oline

shutvik [7]

Answer:

it is refined petrolium used a sfuel for intetnal combustion engines

Explanation:

in simple words petrol

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3 years ago

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3 years ago