The number of species from the following in which the central atom uses sp3 hybrid orbitals in its bonding is ______. \(NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3\)
Correct Answer: 4
Approach Solution - 1
Given Molecules and Their Hybridization:
- NH₃: Ammonia has sp³ hybridization, as the nitrogen atom forms three sigma bonds with hydrogen atoms and has one lone pair, which leads to a tetrahedral geometry.
- SO₂: Sulfur dioxide exhibits sp² hybridization. The sulfur atom forms two sigma bonds with oxygen atoms and has one lone pair, resulting in a bent molecular shape.
- SiO₂: Silicon dioxide also exhibits sp³ hybridization. The silicon atom forms four bonds with oxygen atoms, and its structure adopts a tetrahedral geometry.
- BeCl₂: Beryllium chloride has sp hybridization. The beryllium atom forms two sigma bonds with chlorine atoms in a linear arrangement.
- CO₂: Carbon dioxide has sp hybridization. The carbon atom forms two double bonds with two oxygen atoms in a linear structure.
- H₂O: Water exhibits sp³ hybridization. The oxygen atom forms two sigma bonds with hydrogen atoms and has two lone pairs, giving the molecule a bent structure.
- CH₄: Methane has sp³ hybridization. The carbon atom forms four sigma bonds with hydrogen atoms in a tetrahedral arrangement.
- BF₃: Boron trifluoride exhibits sp² hybridization. The boron atom forms three sigma bonds with fluorine atoms in a trigonal planar geometry.
Correct Answer: The correct answer is 4.
Approach Solution -2
• \(NH_3\): The nitrogen atom is surrounded by three bonded atoms and one lone pair, resulting in an sp3 hybridization due to the tetrahedral arrangement.
• \(SO_2\): Sulfur in \(SO_2\) is sp2 hybridized because it forms two sigma bonds and has one lone pair, giving a bent structure.
• \(SiO_2\): Each silicon atom forms four sigma bonds with oxygen atoms. However, due to its extended lattice structure, we consider the local bonding, indicating sp3 hybridization for the central Si atom.
• \(BeCl_2\): The beryllium atom is sp hybridized as it forms two sigma bonds with chlorine atoms, leading to a linear geometry.
• \(CO_2\): Carbon in \(CO_2\) is sp hybridized since it forms two sigma bonds with oxygen atoms, resulting in a linear structure.
• \(H_2O\): The oxygen atom has two sigma bonds and two lone pairs, leading to sp3 hybridization, resulting in a bent structure.
• \(CH_4\): Carbon in \(CH_4\) is sp3 hybridized as it forms four sigma bonds, resulting in a tetrahedral geometry.
• \(BF_3\): Boron in \(BF_3\) is sp2 hybridized, as it forms three sigma bonds with fluorine atoms, resulting in a planar triangular structure.
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Chemical bonding and molecular structure Questions
- Statement I: Since fluorine is more electronegative than nitrogen, the net dipole moment of NF3 is greater than NH3.
Statement II: In NH3, the orbital dipole due to the lone pair and the dipole moment of NH bonds are in opposite directions, but in NF3, the orbital dipole due to the lone pair and the dipole moments of N-F bonds are in the same direction.
In light of the above statements, choose the most appropriate from the options given below: - What is the structure of Mn2(CO)10?
- Consider the following two statements Statement I: \(π\) \(2p\) bonding molecular orbital has low electron density above & below internuclear axis Statement II: \(π\times2p\) antibonding molecular orbital has only one nodal plane
- Consider the following two statements Statement I: \(SiO_2\) and GeO₂\(GeO_2\) are acidic, \(SnO\) and \(PbO\) are amphoteric Statement II: Allotropes of carbon are formed due to catenation and \(dπ - ρπ\) bond
- Number of compounds with one lone pair of electrons on central atom amongst following is
\(CIF_3, XeO_3, BrF_5, XeF_4, O_3, NH_3\),\( H_2O\)
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,