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2 years ago

The sp3 hybridization has what percent s character and what percent p character respectively?

Chemistry

1 answer:

Salsk061 [2.6K]2 years ago

8 0

Answer:

The sp³ orbital hybridization has 25% s-character and 75% p-character.

Explanation:

Hybridization refers to the process of mixing of a given number atomic orbitals to form equal number of new hybrid orbitals. The three types of hybrid orbitals are sp³, sp² and sp.

The four sp³ hybridized orbitals are formed by the mixing of one s orbital and three p orbitals.

Thus s-character in a sp³ hybridized orbital = (1÷4)×100 = 25%

and p-character in a sp³ hybridized orbital = (3÷4)×100 = 75%

Therefore, the sp³ orbital hybridization has 25% s-character and 75% p-character.

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3 years ago

In the formula for the compound XCl4, the X could represent (1) C (3) Mg (2) Η (4) Zn

Anvisha [2.4K]

Correct Answer: option 1 i.e. C

Reason:
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3 years ago

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marissa [1.9K]

Answer:

Throughout the explanations section below you will find a description of the question.

Explanation:

(1)

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(2)

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3 years ago

This group has characteristics including: brittleness, insulators/poor conductors, and are often found as gases.

bagirrra123 [75]

Answer:

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Explanation:

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3 years ago

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tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. $K(s)+O_2(g)\rightarrow KO_2(s)$

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$

Answer: A. $K(s)+O_2(g)\rightarrow KO_2(s)$ : decreases

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g)$ : decreases

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ : no change

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction $K(s)+O_2(g)\rightarrow KO_2(s)$ , as gaseous reactant is changed to solid product, entropy decreases.

In reaction $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

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3 years ago