An electron-deficient compound is one in which the central atom has fewer than eight electrons in its valence shell, meaning it does not achieve a stable octet. Such compounds can act as Lewis acids.

Let's analyze each option:

• A) SF₆: Sulfur (S) is the central atom. It forms 6 bonds with fluorine atoms. Sulfur has 6 valence electrons, and each fluorine contributes 1 electron to a bond. So, sulfur has $6\times 2=12$ electrons in its valence shell. This is an expanded octet, not electron-deficient.
• B) PCl₅: Phosphorus (P) is the central atom. It forms 5 bonds with chlorine atoms. Phosphorus has 5 valence electrons, and each chlorine contributes 1 electron to a bond. So, phosphorus has $5\times 2=10$ electrons in its valence shell. This is an expanded octet, not electron-deficient.
• C) BCl₃: Boron (B) is the central atom. It forms 3 bonds with chlorine atoms. Boron has 3 valence electrons, and each chlorine contributes 1 electron to a bond. So, boron has $3\times 2=6$ electrons in its valence shell. Since 6 is less than 8, boron does not achieve an octet, making BCl₃ an electron-deficient compound.
• D) ICl: Iodine (I) is the central atom. It forms 1 bond with a chlorine atom. Iodine has 7 valence electrons. After forming one bond, it has $1\times 2=2$ electrons in the bond and 3 lone pairs ($3\times 2=6$ electrons). So, iodine has $2+6=8$ electrons in its valence shell (an octet). It is not electron-deficient.

Therefore, BCl₃ is the electron-deficient compound.

Correct Answer: (C)