Question

2. [6 points By overlapping the following orbitals on neighboring atoms, determine whether o-bonds, t- bonds, or no bonds are formed. (a) (2s, 2s) where the internuclear distance lies along the z-axis. Answer: (b) (2px, 2px) where the internuclear distance lies along the z-axis. Answer: (c) (2px 2px) where the internuclear distance lies along the x-axis. Answer: (d) (2p. 2p,) where the internuclear distance lies along the Z-axis. Answer: (e) (25, 2p.) where the internuclear distance lies along the Z-axis. Answer: (1) (2s, 2px) where the internuclear distance lies along the z-axis. Answer: (a) (b) O O OO 205 28 2s Atom A Atom B 2p.0 2px U2px Atom B Atom A Atom A Atom B (d) a (e) 2px/ 2p 22 Atom A Atom B Atom A Atom B Atom A Atom B

Answer 1

Answer:-

As we know that sigma ($\sigma$) bond is formed by the head to head (head on) overlapping of half filled atomic orbitals along the inter-nuclear axis (i.e the line passing through the centers of nuclei of the   atom) . sigma ($\sigma$​​​​​​​) bond is a strong bond.

Pi ($\pi$) bond is formed by the sidewise (lateral) overlapping of half-filled atomic orbitals in a direction perpendicular to the inter-nuclear axis. Pi ($\pi$) bond is weak bond.

No bond formed if positive lobe of one atomic orbital overlaps with positive lobe of other atomic orbital is equal to the negative lobe of one atomic orbital overlaps with negative lobe of other atomic orbital because in such condition net overlapping is zero.
therefore

(a) - (2s, 2s) where the inter-nuclear distance lies along the z-axis. ; Answer = $\sigma$ - bond.

because there is head to head (head on) overlapping of half filled atomic orbitals along the inter-nuclear axis.

(b) - (2p_x, 2p_x) where the inter-nuclear distance lies along the z-axis. ;   Answer = $\pi$ - bond.

because there is sidewise (lateral) overlapping of half-filled atomic orbitals in a direction perpendicular to the inter-nuclear axis.

(c) - (2p_x, 2p_x) where the inter-nuclear distance lies along the x-axis. ;   Answer = $\sigma$ - bond.

because there is head to head (head on) overlapping of half filled atomic orbitals along the inter-nuclear axis.

(d) - (2p_x, 2p_z) where the inter-nuclear distance lies along the z-axis. ;   Answer = No bond.

because there is positive lobe of one atomic orbital overlaps with positive lobe of other atomic orbital is equal to the negative lobe of one atomic orbital overlaps with negative lobe of other atomic orbital so in such condition net overlapping is zero.

(e) - (2s, 2p_z) where the inter-nuclear distance lies along the z-axis. ;   Answer = $\sigma$ - bond.

because there is head to head (head on) overlapping of half filled atomic orbitals along the inter-nuclear axis.

(f) - (2s, 2p_x) where the inter-nuclear distance lies along the z-axis. ;   Answer = No bond.

because there is positive lobe of one atomic orbital overlaps with positive lobe of other atomic orbital is equal to the negative lobe of one atomic orbital overlaps with negative lobe of other atomic orbital so in such condition net overlapping is zero.