Bond arrangements in carbon allotropes and SiO₂

Steps:

• Identify X: Graphite has sp² hybridized carbon with trigonal planar arrangement (three sigma bonds, one delocalized pi bond).
• Identify Y: Diamond has sp³ hybridized carbon with tetrahedral arrangement (four sigma bonds).
• Identify Z: SiO₂ has sp³ hybridized silicon with tetrahedral arrangement (four sigma bonds to oxygen).
• Compare: Y and Z share tetrahedral geometry; X differs as trigonal planar.

Why C is correct:

• Y and Z both exhibit tetrahedral coordination per VSEPR theory, with bond angles near 109.5°.

Why the others are wrong:

• A: X (trigonal planar) differs from Y (tetrahedral).
• B: X (trigonal planar) differs from Z (tetrahedral).
• D: X's trigonal planar arrangement is not tetrahedral like Y and Z.

Final answer: C