• \(NH_3\): 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.
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.