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On hybridisation of one s and three p-or...

On hybridisation of one s and three p-orbitals, we get

A

four orbitals with tetrahedral orientation

B

three orbitals with trigonal orientation

C

two orbitals with linear orientation

D

two orbitals with perpendicular orientation.

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To solve the question regarding the hybridization of one s and three p orbitals, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Orbitals Involved**: - We have one s orbital and three p orbitals involved in the hybridization process. 2. **Determine the Type of Hybridization**: - When one s orbital and three p orbitals combine, they undergo hybridization to form new hybrid orbitals. The combination of one s and three p orbitals leads to the formation of four equivalent hybrid orbitals. 3. **Name the Hybridization**: - The hybridization formed from one s and three p orbitals is called **sp³ hybridization**. This is because the 's' contributes one orbital and the three 'p' orbitals contribute three, totaling four orbitals. 4. **Shape and Orientation of Hybrid Orbitals**: - The four sp³ hybrid orbitals are arranged in a tetrahedral geometry. This arrangement minimizes the repulsion between the orbitals, resulting in a stable molecular structure. 5. **Example of sp³ Hybridization**: - A common example of a molecule with sp³ hybridization is methane (CH₄). In methane, the carbon atom undergoes sp³ hybridization to form four equivalent sp³ hybrid orbitals, which are oriented tetrahedrally around the carbon atom. 6. **Conclusion**: - Therefore, on hybridization of one s and three p orbitals, we get four sp³ hybridized orbitals with a tetrahedral orientation. ### Final Answer: On hybridization of one s and three p orbitals, we get **four sp³ hybridized orbitals with tetrahedral orientation**. ---

To solve the question regarding the hybridization of one s and three p orbitals, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Orbitals Involved**: - We have one s orbital and three p orbitals involved in the hybridization process. 2. **Determine the Type of Hybridization**: ...
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According to hybridisation theory, the % s-character in sp, sp^(2) and sp^(3) hybrid orbitals is 50, 33.3 and 25 respectively, but this is not true for all the species. When theta is the bond angle between equivalent hybrid orbitals then % s and p-character in hybrid orbitals (when only s and p-orbitals are involved in hybridisation) can be calculated by the following formula : costheta=(S)/(S-1)=(P-1)/(P) Q. Correct order of P-P bond length in the following compound is :

According to hybridisation theory, the % s-character in sp, sp^(2) and sp^(3) hybrid orbitals is 50, 33.3 and 25 respectively, but this is not true for all the species. When theta is the bond angle between equivalent hybrid orbitals then % s and p-character in hybrid orbitals (when only s and p-orbitals are involved in hybridisation) can be calculated by the following formula : costheta=(S)/(S-1)=(P-1)/(P) Q. Correct order of P-P bond length in the following compound is :

According to hybridisation theory, the % s-character in sp, sp^(2) and sp^(3) hybrid orbitals is 50, 33.3 and 25 respectively, but this is not true for all the species. When theta is the bond angle between equivalent hybrid orbitals then % s and p-character in hybrid orbitals (when only s and p-orbitals are involved in hybridisation) can be calculated by the following formula : costheta=(S)/(S-1)=(P-1)/(P) Q. Two elements X and Y combined together to form a covalent compound. If % p-character is found to be 80% in a orbital then the hybridised state of central atom X for the orbital is :

According to hybridisation theory, the % s-character in sp, sp^(2) and sp^(3) hybrid orbitals is 50, 33.3 and 25 respectively, but this is not true for all the species. When theta is the bond angle between equivalent hybrid orbitals then % s and p-character in hybrid orbitals (when only s and p-orbitals are involved in hybridisation) can be calculated by the following formula : costheta=(S)/(S-1)=(P-1)/(P) Q. Smallest Ohat(S)O bond angle is found in :

When hybridisation involving d-orbitals are considered then all the five d-orbitals are not degenerate, rather d_(x^(2)-y^(2)),d_(s^(2)) and d_(xy),d_(zx) form two different sets of orbitals and orbitals of appropriate set is involved in the hybridisation. Which of the following orbitals can not undergo hybridisation amongst themselves. (I) 3d,4s " " (II) 3d,4d (III) 3d,4s & 4p " " (IV) 3s,3p & 4s

NCERT FINGERTIPS ENGLISH-CHEMICAL BONDING & MOLECULAR STRUCTURE-Assertion And Reason
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  2. Assertion : In the formation of a molecule , only the outer shell ele...

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  3. Assertion : In the formation of water molecule, both hydrogen and oxyg...

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  4. Assertion : In Liwis structures of NF(3) and CO(3)^(2-), nitrogen and...

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  5. Assertion : PF(5),SF(6)and H(2)SO(4) are the examples of expanded octe...

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  6. Assertion : Octet rule is based upon the chemical inertness of noble g...

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  7. Assertion : Sodium chloride (NaCl) is a stable ionic solid. Reason ...

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  8. Assertion : F(2)and O(2)^(2-)have bond order 1 while N(2),CO and NO^(+...

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  9. Assertion : The experimentally determined carbon to oxygen bond length...

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  10. Assertion : The dipole moment in case of BeF(2) is zero. Reason : T...

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  11. Assertion : Dipole moment of NH(3) is greater than that of NF(3). Re...

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  12. Assertion : Among alkaline earth metals, Be predominantly forms covale...

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  13. Assertion : In NH(3), N "is " sp^(3) hybridised but bond angle is 107^...

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  14. Ionic bonds are non-directional while covalent bonds are directional.

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  15. Assertion : O(2) molecule is diamagnetic while C(2) molecule is parama...

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  16. Assertion : Boiling point of p-nitrophenol is greater than that of o-n...

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