Question

Number 1-6

1. What is the molecular geometry of SBr4? A) seesaw B) square planar C) square pyramidal D) tetrahedral 2. Using the VSEPR model, the molecular geometry of the central atom in CF4 is A) linear B) trigonal planar C) tetrahedral D) bent E) trigonal pyramidal 3. Using the VSEPR model, the molecular geometry of the central atom in NC13 is A) linear B) trigonal planar C) tetrahedral D) bent E) trigonal pyramidal 4. Using the VSEPR model, the molecular geometry of the central atom in BF3 is linear B) trigonal planar tetrahedral D) bent E) trigonal pyramidal i r er al B) 2 C) 3 D) 4 E) 5 6. The molecular structure of AsCls is A trigonal bipyramidal B) square pyramidal C) distorted tetrahedral D) octahedral E) none of these

Answer 1

S, 1s^2, 2s^2, 2p^6, 3s^2, 3p^4 Br = [10]ns^2 np^5 sp^3 d hybridization sp63 cl hybridization with lp = 1 cf_4 c_6 = 1s^2, 2s^2, 2sp^2 NCl_3 N_7 = 1s^2, 2s^2, 2p^3 NCl_3 sp^3 hybridization lp trigonal pyramidal BF_3_5 B = 1s^2, 2s^2 2p^1 sp^2 hybridization therefore trigonal planar shape octahedral molecular have six atoms bonded to the central atom in and no lone pour electron Hence stone number is Answer is A Ascl_3 As is included in nitrogen family hence general electronic configuration 2ns^2 np^3 sp63 dlg hybridization molecular geometry is trigonal bi pyramidal

Answer 2

A Lewis structure is used to determine the electronic geometry as well as molecular geometry of a molecule A Lewis structure is a skeletal structure with all the elements and non-bonding electrons shown. Two electrons indicate one electron pair The table for hybridization with respect to electronic geometry using different combinations of bonds and lone pairs of electrons present on central atom is below: Number of atom/group attached to par of electrons Number of lone Hybridization of the central atom Electron geometry on central atom central atom Linear Trigonal planar Trigonal planar Tetrahedral Tetrahedral Tetrahedral Trigonal bipyramidal Trigonal bipyramidal Trigonal bipyramidal Trigonal bipyramidal Sp'd Sp d Sp'd Sp'd Sp'd Sp d 4 Octahedral Answer 1- In compound SB, S İs the central atom. The valance shell electronic configuration of s is 3s, 3p It has 6 valance electron which are involved in bonding. Each S electron is making single bond with four Br atoms and one pair of electron is left. Therefore, it has 4 bond pair and 1 lone pair Hence the hybridization of the compound according to the table mentioned above is sp d and the electron geometry is trigonal bipyramidal as shown below: The molecular geometry of the compound is see saw as seen in the figure. Hence, the correct option is A.

Answer 2 compound CF4, Cis the central atom. The It has 4 valance electron which are involved in bonding. Each C electron is making single bond with four F atoms. Therefore, it has 4 bond pair and no lone pair. Hence the hybridization of the compound according to the table mentioned above is sp and the electron geometry is tetrahedral as shown below: The molecular geometry of the compound also tetrahedral as seen in the figure. Hence, the correct option is C.