No, a key and IV give very different security properties, and you can't compensate a weak IV with a strong key, or the other way around.
First of all, a small misconception: a 16-bit number can have $2^{16}$ possible values, and combining them gives $2^{16} \cdot 2^{16} = 2^{32}$ possible values, or a 32-bit number, not a 256-bit number.
An IV or initialisation vector is used to prevent chosen-plaintext attacks. Without an IV, encrypting a specific plaintext with a specific key will always yield the exact same ciphertext. An attacker can use this to its advantage by asking an encryption oracle to encrypt a plaintext that the attacker guesses. If the ciphertext matches the original encrypted message, the attacker knows that the chosen plaintext is the text that was originally sent.
An IV will change the ciphertext each time a message is encrypted, even if the key and plaintext are exactly the same. It is important that the IV is unpredictable before encrypting the text, but it does not have to be a secret after the text is encrypted. Even if the attacker knows the IV for a specific message, when the attacker asks the encryption oracle to encrypt a chosen plaintext, the IV will be different, so the attacker can't verify that the original message was the same as the guessed plaintext.
No matter how strong your encryption key is, it won't protect against this kind of attack.
Similarly, a long IV won't protect against a weak key. Since the recipient of the message needs to know the IV, it is usually send in cleartext along with the encrypted message. Now an attacker only has to guess the key, so if the key is weak and the attacker guesses correctly, s/he can actually decrypt all messages that use that key.